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 1818 phones from the manufacturer had a mean range of 10201020 feet with a standard deviation of 2525 feet. A sample of 1313 similar phones from its competitor had a mean range of 10101010 feet with a standard deviation of 2929 feet. Do the results support the manufacturer's claim? Let μ1μ1 be the true mean range of the manufacturer's cordless telephone and μ2μ2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.01α=0.01 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.

Compute the value of the t test statistic. Round your answer to three decimal places.

Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places.

State the test's conclusion.

Answer 1

given data and necessary calculation are:-

the pooled standard deviation be:-

a).hypothesis:-

[ claim ]

b).the test statistic be:-

c).degrees of freedom = (18+13-2) = 29

t critical value for df=29,alpha=0.01, right tailed test be:-

[ using t distribution table ]

decision rule :-

reject the null hypothesis if,

d)conclusion:-

we fail to reject the null hypothesis and conclude that there is not enough to claim that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor .

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