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 19 phones from the manufacturer had a mean range of 1160 feet with a standard deviation of 32 feet. A sample of 11 similar phones from its competitor had a mean range of 1130 feet with a standard deviation of 30 feet. Required:
Do the results support the manufacturer's claim?

Complete question is;

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 19 phones from the manufacturer had a mean range of 1160 feet with a standard deviation of 32 feet. A sample of 11 similar phones from its competitor had a mean range of 1130 feet with a standard deviation of 30 feet. Required:

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 = 0.01 for the test. Assume that the population variances are equal and that the two populations are normally distributed

We will fail to reject the null hypothesis as there is no sufficient evidence to support the manufacturers claim.

Step-by-step explanation:

For the first sample, we have;

Mean; x'1 = 1160 ft

standard deviation; σ1 = 32 feet

Sample size; n1 = 19

For the second sample, we have;

Mean; x'2 = 1130 ft

Standard deviation; σ2 = 30 ft

Sample size; n2 = 11

The hypotheses are;

Null Hypothesis; H0; μ1 = μ2

Alternative hypothesis; Ha; μ1 > μ2

The test statistic formula for this is;

z = (x'1 - x'2)/√[(σ1)²/n1) + (σ2)²/n2)]

Plugging in the relevant values, we have;

z = (1160 - 1130)/√[(32)²/19) + (30)²/11)]

z = 2.58

From the z-table attached, we have a p-value = 0.99506

This p-value is more than the significance value of 0.01,thus,we will fail to reject the null hypothesis as there is no sufficient evidence to support the manufacturers claim.