[1] An earlier study claims that U.S. adults spend an average of 114 minutes with their families per day. A recently taken sample of 25 adults showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the time spent by adults with their families has an approximate normal distribution. We wish to test whether the mean time spent currently by all adults with their families is less than 114 minutes a day.
a) Set up the null and alternative hypotheses.
Null Hypothesis (Ho): µ = 114
Alternative Hypothesis (Ha): µ < 114
b) Select the distribution to use. Explain briefly why you selected it.
The distribution selected for this is t distribution because sample size is less than 30.
If the size of a sample is less than 30, in general you can't use a t-distribution unless...
Did you check your work in [4] of Lab #2 with my solution?
c) Using the 2.5% significance level, determine the rejection and non-rejection regions based on your hypotheses in a). State the critical value.
Since this is a lower tailed test so the lower critical value will be with n-1= 25-1=24 degrees of freedom at 0.025 level of significance is -2.0639. So the rejection region is when test statistics value is smaller than the lower critical value and non-rejection region is when test statistics value is bigger than the lower critical value.
d) Calculate the value of the test statistic.
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t Test for Hypothesis of the Mean
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Data
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Null Hypothesis m=
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114
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Level of Significance
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0.025
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Sample Size
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25
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Sample Mean
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109
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Sample Standard Deviation
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11
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Intermediate Calculations
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Standard Error of the Mean
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2.2000
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Degrees of Freedom
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24
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t Test Statistic
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-2.2727
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Lower-Tail Test
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Lower Critical Value
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-2.0639
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p-Value
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0.0161
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Reject the null hypothesis
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Test statistics value is -2.2727.
e) Does the sample information support that the mean time spent currently by all adults with their families is less than 114 minutes a day? Explain your conclusion in words.
Since the test statistics value is smaller than lower critical value so we will reject the null hypothesis. The sample information support that the mean time spent currently by all adults with their families is less than 114 minutes a day.
[2] In recent years, the Town of New Port experienced an arrest rate of 25% for robberies (based on FBI data). The new sheriff complies records showing that among 30 recent robberies, the arrest rate is 30%, so she claims that her arrest rate is greater than the 25% rate in the past. We want to test if there is sufficient evidence to support her claim.
a) Set up the null and alternative hypotheses, and perform the hypothesis test with a significance level of 0.05.
Null Hypothesis (Ho): p = 0.25
Alternative Hypothesis (Ha): p < 0.25
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Z Test of Hypothesis for the Proportion
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Data
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Null Hypothesis p =
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0.25
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Level of Significance
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0.05
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Number of Items of Interest
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9
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Sample Size
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30
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Intermediate Calculations
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Sample Proportion
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0.3
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Standard Error
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0.0791
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Z Test Statistic
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0.6325
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Upper-Tail Test
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Upper Critical Value
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1.6449
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p-Value
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0.2635
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Do not reject the null hypothesis
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This is a binomial situation (arrested or not). But you used a Z-test. Why not a t-test or something else? Have you reviewed your work on [3] of Lab #2?
b) Does the sample information support that her claim that the arrest rate is greater than 25%? Explain your conclusion in words.
Since the test statistics value is smaller than the upper critical value, we will not be able to reject the null hypothesis and conclude that the sample information does not support that her claim that the arrest rate is greater than 25%.
[3] A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $74.50. We shall use a 0.05 significance level to test the claim that a family of four will spend more money, on average, when vacationing at Niagara Falls per day than the average amount for a family of four claimed by the American Automobile Association.
a) Set up the null and alternative hypotheses, and perform the hypothesis test.
Null Hypothesis (Ho): µ = 215.60
Alternative Hypothesis (Ha): µ > 215.60
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Z Test of Hypothesis for the Mean
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Data
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Null Hypothesis m=
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215.6
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Level of Significance
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0.05
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Population Standard Deviation
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74.5
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Sample Size
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64
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Sample Mean
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252.45
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Intermediate Calculations
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Standard Error of the Mean
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9.3125
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Z Test Statistic
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3.9570
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Upper-Tail Test
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Upper Critical Value
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1.6449
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p-Value
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0.0000
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Reject the null hypothesis
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Why is this by a Z-test? You didn't use the Z-distribution in [1] of Lab #2?
Based on the hypothesis test, we will reject the null hypothesis.
b) Based on results of the hypothesis test in part a), does it appear that the population mean amount spent per day by families visiting Niagara Falls is more than the mean amount per day for a family of four reported by the American Automobile Association? Explain.
Since the test statistics value is bigger than the upper critical value so we will reject the null hypothesis and conclude that it appear that the population mean amount spent per day by families visiting Niagara Falls is more than the mean amount per day for a family of four reported by the American Automobile Association.
[4] In the case of Casteneda v. Partida, 1977, it was found that during a period of 11 years in Hilda County, Texas, 870 people were selected for grand jury duty, and 39% of them were Americans of Mexican ancestry. Among the people eligible for grand jury duty, 79.1% were Americans of Mexican ancestry. We shall use a 0.02 significance level to test the claim that the selection process is biased against Americans of Mexican ancestry.
a) Set up the null and alternative hypotheses, and perform the hypothesis test.
Null Hypothesis (Ho): p = 0.791
Alternative Hypothesis (Ha): p ≠ 0.791
Level of significance = 0.02
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Z Test of Hypothesis for the Proportion
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Data
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Null Hypothesis p =
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0.791
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Level of Significance
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0.02
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Number of Items of Interest
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339
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Sample Size
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870
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Intermediate Calculations
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Sample Proportion
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0.389655172
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Standard Error
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0.0138
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Z Test Statistic
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-29.1149
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Two-Tail Test
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Lower Critical Value
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-2.3263
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Upper Critical Value
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2.3263
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p-Value
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0.0000
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Reject the null hypothesis
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This is a binomial situation (American of Mexican ancestry or not). But you used a Z-test. Why not a t-test or something else? Have you reviewed your work on [3] of Lab #2?
Based on the hypothesis test, we will reject the null hypothesis.
b) Does the jury selection system appear to be fair?
Since the test statistics value is smaller than the lower critical value so we will reject the null hypothesis and conclude that jury selection system does not appear to be fair.
But since your setup is for a two-tail test, your finding in (a) is that there is sufficient evidence that p (whatever this represents) is NOT 79.1% (it could be larger or smaller than 79.1%). Thus you need to explain why this leads to your conclusion in (b).
[5] A bakery in my neighborhood produces loaves of bread with "1 pound" written on the label. Weights of randomly selected sampled loaves from today's production were recorded below:
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1.02
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0.97
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0.98
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1.04
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1.02
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1.00
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0.98
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1.03
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1.05
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0.99
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1.02
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1.06
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0.98
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1.01
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0.99
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1.02
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a) If you are to perform a hypothesis test to test the claim on the label that the mean weight of the entire loaves of bread production is more than "1 pound," what distribution are you going to apply and explain why?
We will apply the t distribution because the sample size is less than 30 and we want to test the population mean.
Now I really see that you don't review what you have done with your graded work. This have been the most important issue in the lab #2.
b) If an average weight is less than or equal to "1 pound," then customers think that the bakery is taking advantage of its weight. When customers find out its actual weight, they might certainly complain about it. Use a 0.05 significance level to test the claim that the mean weight of the entire loaves of bread production is more than "1 pound."
Null Hypothesis (Ho): µ = 1
Alternative Hypothesis (Ha): µ > 1
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t Test for Hypothesis of the Mean
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Data
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Null Hypothesis m=
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1
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Level of Significance
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0.05
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Sample Size
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16
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Sample Mean
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1.01
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Sample Standard Deviation
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0.027080128
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Intermediate Calculations
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Standard Error of the Mean
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0.0068
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Degrees of Freedom
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15
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t Test Statistic
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1.4771
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Upper-Tail Test
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Upper Critical Value
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1.7531
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p-Value
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0.0802
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Do not reject the null hypothesis
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Based on the hypothesis test, we will not reject the null hypothesis.
c) Do you think that the bakery is correct in its label claim?
We can see that the test statistics value is smaller than the upper critical value so we will not be able to reject the null hypothesis and conclude that the mean weight of the entire loaves of bread production is not more than "1 pound."
Not correct to state above, instead, state that there is not enough (or insufficient) evidence that the mean weight of the entire loaves of bread production is more than "1 pound."
[6] On January 7, 2000, the Gallup Organization released the results of a poll comparing lifestyles of today with that yesteryear. Then poll results were based in telephone interviews with a randomly selected national sample of 1,031 adults, 18 years and older, conducted December 20-21, 1999. One question asked if the respondent had vacationed for six days or longer within the last 12 months. Suppose that we will attempt to use the poll's results to justify the claim that more than 40 percent of U.S. adults have vacationed for six days or longer within the last 12 months. The poll actually found that 42 percent of the respondents had done so. Would you conclude that more than 40 percent of U.S. adults have vacationed for six days or longer within the last 12 months? Explain.
This is a binomial situation (yes or no). But you used a Z-test. Why not a t-test or something else? Have you reviewed your work on [3] of Lab #2?
Null Hypothesis (Ho): p = 0.40
Alternative Hypothesis (Ha): p > 0.40
Level of significance = 0.05
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Z Test of Hypothesis for the Proportion
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Data
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Null Hypothesis p =
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0.4
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Level of Significance
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0.05
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Number of Items of Interest
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433
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Sample Size
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1031
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Intermediate Calculations
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Sample Proportion
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0.419980601
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Standard Error
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0.0153
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Z Test Statistic
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1.3096
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Upper-Tail Test
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Upper Critical Value
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1.6449
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p-Value
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0.0952
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Do not reject the null hypothesis
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Since the test statistics value is smaller than the upper critical value so we will not be able to reject the null hypothesis and conclude that not more than 40 percent of U.S. adults have vacationed for six days or longer within the last 12 months.
Again the correct way to state your finding is that there is insufficient evidence that more than 40 percent of U.S. adults have vacationed for six days or longer within the last 12 months. This is because you test on the claim (null hypothesis) using sample data, not the entire population. It won't be 100% sure about the claim. So you can conclude that there is sufficient evidence or insufficient evidence that ..., instead of stating that you can conclude that ...