--%>

Variance and standard error

A hospital treated 412 skin cancer patients over a year. Of these, 197 were female.

Give the point estimate of the proportion of females seeking treatment for skin cancer.

Give estimates of the variance and standard error of the point estimate.

Give a 95% confidence interval for the population proportion of females seeing treatment for skin cancer.

Use an appropriate test to determine whether this dataset provides statistically significant evidence that males are more likely to seek treatment for skin cancer.

E

Expert

Verified

Given:

n: Total number of patients = 412
x1: Number of females in the sample seeking cancer treatment = 197
x2: Number of males in the sample seeking cancer treatment = 412-197 = 215
   
Define:

p1: sample proportion of females seeking cancer treatment.

p1= x1/n = 197/412 = 0.4782

p1: sample proportion of males seeking cancer treatment.

p2 = x2/n = 215/412 = 0.5218

We know that the sample proportion is an unbiased estimator of population proportion, hence the proportion of females seeking treatment for skin cancer is

p ^= p1=0.4782

In case of proportions, the population variance is estimated by:

2061_stats2.jpg

The standard error is nothing but the square root of variance
Hence,

1909_stats3.jpg


The 95% confidence interval for the population proportion of females seeing treatment for skin cancer is given by:

1631_stats4.jpg

Where, ZC (Critical value) = 1.96
Hence,

1673_stats5.jpg

This is the required confidence interval.

Now we are supposed to test whether males are more likely to seek treatment for skin cancer.

Null hypothesis:

H0: There is no significant difference in number of cancer patients due to according to gender
H0: P1 = P2

Alternative hypothesis:

Ha: Males are more likely to seek treatment for skin cancer.
Ha: P1 < P2

α (level of significance) = 0.05         One tailed test
Zα (Critical value) = -1.64

Assumptions:

The two samples come from independent population.
Population is normally distributed.

Test Statistic:

72_stats6.jpg

Where,
 
P = 1/2

Q= 1/2

Hence Z = - 1.2541

 P value = P (Z < Z observed)
             = P (Z < -1.2541 )
             = 0.1049

Decision Rule:

Reject H0 if P value is less than the level of significance.

Decision:

Since observed value (-1.2541) > critical value (-1.64) and P value (observed level of significance) = 0.1049 is greater than α (level of significance) = .05, we fail to reject H0.

Conclusion:

There is no significant difference in number of cancer patients due to according to gender.

   Related Questions in Basic Statistics

  • Q : How to solve statistics assignment in

    How to solve staistics assignment, i need some help in solving stats assignment on AVOVA based problems. Could you help in solving this?

  • Q : Write out the null hypothesis 1.

    1. (AAC/ACA c9q1).  For each of the following studies, decide whether you can reject the null hypothesis that the groups come from identical populations. Use the alpha = .05 level.1a.

  • Q : Define Service Demand Law

    Service Demand Law:• Dk = SKVK, Average time spent by a typical request obtaining service from resource k• DK = (ρk/X

  • Q : Hw An experiment is conducted in which

    An experiment is conducted in which 60 participants each fill out a personality test, but not according to the way they see themselves. Instead, 20 are randomly assigned to fill it out according to the way they think a parent sees them (i.e. how a parent would fill it out to describe the participant

  • Q : Hypothesis homework A sample of 9 days

    A sample of 9 days over the past six months showed that a clinic treated the following numbers of patients: 24, 26, 21, 17, 16, 23, 27, 18, and 25. If the number of patients seen per day is normally distributed, would an analysis of these sample data provide evidence that the variance in the numbe

  • Q : Calculate the p- value Medical tests

    Medical tests were conducted to learn about drug-resistant tuberculosis. Of 284 cases tested in New Jersey, 18 were found to be drug- resistant. Of 536 cases tested in Texas, 10 were found to be drugresistant. Do these data indicate that New Jersey has a statisti

  • Q : Homework help on Human memory & SPSS

    Effect of Scopolamine on Human Memory: A Completely Randomized Three Treamtent Design (N = 28) Scopolamine is a sedative used to induce sle

  • Q : Compute two sample standard deviations

    Consider the following data for two independent random samples taken from two normal populations. Sample 1 14 26 20 16 14 18 Sample 2 18 16 8 12 16 14 a) Com

  • Q : Cumulative Frequency and Relative

    Explain differences between Cumulative Frequency and Relative Frequency?

  • Q : MANOVA and Reflection Activity

    Activity 10:   MANOVA and Reflection   4Comparison of Multiple Outcome Variables This activity introduces you to a very common technique - MANOVA. MANOVA is simply an extension of an ANOV

  • ©TutorsGlobe All rights reserved 2022-2023.