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 8 phones from the manufacturer had a mean range of 1300 feet with a standard deviation of 43 feet. A sample of 14 similar phones from its competitor had a mean range of 1280 feet with a standard deviation of 36 feet. Do the results support the manufacturer's claim? Let μ1μ1 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=0.1 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 test statistic. Round your answer to two decimal places.

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

Step 4 of 4: Make the decision for the hypothesis test.

Answer #1

1)

Below are the null and alternative Hypothesis,

Null Hypothesis, H0: μ1 = μ2

Alternative Hypothesis, Ha: μ1 > μ2

b)

Pooled Variance

sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 +
1/n2))

sp = sqrt((((8 - 1)*43^2 + (14 - 1)*36^2)/(8 + 14 - 2))*(1/8 +
1/14))

sp = 17.1053

Test statistic,

t = (x1bar - x2bar)/sp

t = (1300 - 1280)/17.1053

t = 1.17

c)

Rejection Region

This is right tailed test, for α = 0.1 and df = n1 + n2 - 2 =
20

Critical value of t is 1.33.

Hence reject H0 if t > 1.33

d)

Fail to reject the null hypothesis

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