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 9 phones from the manufacturer had a mean range of 1120 feet with a standard deviation of 383 feet. A sample of 16 phones from the manufacturer had a mean range of 1050 feet with a standard deviation of 23 feet. A sample of 12 similar phones from its competitor had a mean range of 1020 feet with a standard deviation of 41 feet. 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.
a. State the null and alternative hypotheses for the test.
b. Compute the value of the t test statistic. Round your answer to three decimal places.
c. Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.
d. State the test's conclusion.

Answer 1

Answer:

Step-by-step explanation:

Hello!

The manufacturer hypothesizes that the calling range of its 900-MHz cordless telephone is greater than its leading competitor, if:

X₁: Calling range of the 900-MHz cordless telephone of the manufacturer. (feet)

X₂: Calling range of the 900-MHz cordless telephone of the competitor. (feet)

Sample 1

n₁= 16 phones

X[bar]₁= 1050 feet

S₁= 23 feet

Sample 2

n₂= 12 phones

X[bar]₂= 1020 feet

S₂= 41 feet

The parameters of interest are the population means of the calling range of the phones if the hypothesis of the manufacturer is true, then the statistical hypotheses are:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.01

The populations are normally distributed, independent and their population variances are equal, the statistic to use is the pooled t-test with a pooled standard deviation:

$t= \frac{(X[bar]_1-X[bar_2])-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1}+ \frac{1}{n_2} } }$

$Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$

$Sa^2= \frac{(16-1)(23^2)+(12-1)(41)^2}{16+12-2}= 1016.38$

$Sa= 31.88$

$t_{H_0}= \frac{(1050-1020)-0}{31.88\sqrt{\frac{1}{16}+ \frac{1}{12} } }= 13.91$

This pooled t-test has n₁+n₂-2 degrees of freedom and the test is one-tailed to the right, so you have only one critical value:

$t_{n_1+n_2-2;1-\alpha }= t_{26;0.99}= 2.479$

If the t value is equal or greater than 2.479 you decide to reject the null hypothesis.

If the t value is less than 2.479 you decide to not reject the null hypothesis.

Since the calculated t value is greater than the critical value, you have to reject the null hypothesis.

At 1% significance level, there is significant evidence to reject the null hypothesis, so you can say that the population mean of the call range of the cordless phones of the manufacturer is larger than the call range of the cordless phones made by the competition.

I hope you have a SUPER day!