Question

A company surveyed 2500 men where 1225 of the men identified themselves as the primary grocery shopper in their household. ​a) Estimate the percentage of all males who identify themselves as the primary grocery shopper. Use a 98​% confidence interval. Check the conditions first. ​b) A grocery store owner believed that only 44​% of men are the primary grocery shopper for their​ family, and targets his advertising accordingly. He wishes to conduct a hypothesis test to see if the fraction is in fact higher than 44​%. What does your confidence interval​ indicate? Explain. ​c) What is the level of significance of this​ test? Explain.

Answer 1

Answer:

a) The 98% confidence interval for the population proportion of males who identify themselves as the primary grocery shopper is (0.457, 0.503).

b) The null hypothesis that the proportion is equal or less than 44% will be rejected at this level of significance.

c) The level of significance is 2%.

This is the expected probability that the sample proportion fall outside the acceptance region due to pure chance, if the null hypothesis is true.

Step-by-step explanation:

a) The conditions are:

- randomization: it is assumed to be a random sample.

- independence: the sample responses are independent of each other.

- 10% condition: we don't know the size of the population, but the sample is assumed to be less than 10% of the population.

- sample size: the sample size is bigger than 30, so it is satisfied.

We have to calculate a 98% confidence interval for the proportion.

The sample proportion is p=0.48.

$p=X/n=1225/2550=0.48$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.48*0.52}{2550}}\\\\\\ \sigma_p=\sqrt{0.0001}=0.0099$

The critical z-value for a 98% confidence interval is z=2.326.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.326 \cdot 0.0099=0.023$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.48-0.023=0.457\\\\UL=p+z \cdot \sigma_p = 0.48+0.023=0.503$

The 98% confidence interval for the population proportion of males who identify themselves as the primary grocery shopper is (0.457, 0.503).

b) Our confidence interval tells us that there is 98% confidence that the true proportion o males who identify themselves as the primary grocery shopper is within 0.457 and 0.503.

The propotion 0.44 is outside of this interval, so if the claim is that the proportion is higher than 44%, the null hypothesis will state that the true proprotion is equal or less than 0.44.

With the information we have, at this level of significance (2%), the null hypothesis will be rejected.

c) The level of significance that corresponds to the 98% confidence interval is α=2%.

This is the expected probability that the sample proportion fall outside the acceptance region (which matches with the confidence interval) due to pure chance, if the null hypothesis is true.