You wish to test the following claim (Ha) at a significance level of α=0.05. Ho:p1=p2 Ha:p1<p2You obtain 63 successes in a sample of size n1=797 from the first population. You obtain 30 successes in a sample of size n2=288from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.What is the test statistic for this sample? (Report answer accurate to two decimal places.)test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = The p-value is...less than (or equal to) αgreater than αThis test statistic leads to a decision to...reject the nullaccept the nullfail to reject the nullAs such, the final conclusion is that...There is sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.There is not sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.The sample data support the claim that the first population proportion is less than the second population proportion.There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.

Denote p1, p2 as the true population proportions of first and second populations, respectively. Null hypothesis: H0: p1 = p2. Alternative hypothesis: Ha: p1 < p2. (claim)It is given that, x1 = 63. n1 = 797, x2 = 30 and n2 = 288. The sample proportions are: p̂1 ≈ 0.079046 (= 63/797). p̂2 ≈ 0.104167 (= 30/288). The pooled proportion is: p = (n1p̂1 + n2 p̂2 ) / (n1 + n2). [Here, n1 p̂1 = x1 and n2 p̂2 = x2.] = [63 + 30] / [797 + 288] ≈ 0.085714285. The test statistic is: z = (p̂1 – p̂2 )/ √ [p (1 – p) (1/n1 + 1/n2)] = (0.079046 – 0.104167) / √ [(0.085714285) (1 – 0.085714285) (1/797 + 1/288)] = –0.025121/0.019246733 ≈ –1.31. Thus, the test statistic is –1.31. The level of significance is, α = 0.05 or 5%. Using the Excel formula =NORM.S.DIST(-1.31,TRUE) the p-value is 0.0951. The decision rule regarding a hypothesis testing problem using the p-value is: If p-value ≤ α, reject the null hypothesis H0. Otherwise fail to reject H0. Since p-value (= 0.0951) > α (= 0.05), fail to reject H0. The p-value is greater than α. This test statistic leads to a decision to fail to reject the null. There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.

Answered: You wish to test the following claim…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

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You wish to test the following claim (Ha) at a significance level of α=0.05.

Ho:p1=p2
Ha:p1<p2

You obtain 63 successes in a sample of size n1=797 from the first population. You obtain 30 successes in a sample of size n2=288from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

What is the test statistic for this sample? (Report answer accurate to two decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

less than (or equal to) α
greater than α

This test statistic leads to a decision to...

reject the null
accept the null
fail to reject the null

As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.
There is not sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.
The sample data support the claim that the first population proportion is less than the second population proportion.
There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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