Question

A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 15phones from the manufacturer had a mean range of 1250 feet with a standard deviation of 23 feet. A sample of 6 similar phones from its competitor had a mean range of 1200 feet with a standard deviation of 44feet. Do the results support the manufacturer's claim? Let μ1 be the true mean range of the manufacturer's cordless telephone and μ2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion. (Reject or Fail to Reject Null hypothesis)

Answer 1

The null and alternative hypothesis are:

; i.e., the true mean calling range of manufacturer cordless telephone and its leading competitor are not different than each other, Or we can say its equal.

; i.e., the true mean calling range of manufacturer cordless telephone is greater than that of its leading competitor.

We need to test this hypothesis at a given significance level of .

Since , it is given in the question that the population variances are equal and that the two populations are normally distributed, so the appropriate test to test this hypothesis is Two-sample t test for Equal variance.

Given: We have given the data for sample phones for Manufacturer and competitor.

	Manufacturer	Competitor
sample mean
sample standard deviation
sample size

Test-statistic:

where,

Degree of freedom:

Using the given information we now calculate the test-statistic.

So, the test statistic is calculated as

P-value:

We need the p-value in order to make the decision whether to reject or fail to reject null hypothesis. Since it is a Right tailed hypothesis, and we have calculated the test-statistic as and the degrees of freedom is found out to be as

Since,

At the sample data provide enough evidence to support the alternative hypothesis H₁. Hence, the data supports the Manufacturer's claim that the true mean calling range of manufacturer cordless telephone is grater than that of its leading competitor.

In other words, at we reject null hypothesis, and conclude that the data provide enough evidence to support the alternative hypothesis, i.e., , so, based on the sample the Manufacturer's claim that his true mean calling range of cordless phone is greater than that of its leading competitor.