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 6 phones from the manufacturer had a mean range of 1300 feet with a standard deviation of 20 feet. A sample of 12 similar phones from its competitor had a mean range of 1290 feet with a standard deviation of 42 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=0.05

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.

Answer 1

Step 1 of 4 : The null and alternative hypothesis is ,

( Claim)

The test is right tailed test.

Step 2 of 4 : Since , the population variances are equal and unknown.

Therefore , use t-distribution.

Now , df=degrees of frrdom=n1+n2-2=6+12-2=16

The critical value is ,

; From t-table

Step 3 of 4 : Now , the pooled estimate is ,

The test statistic is ,

Step 4 of 4 : Decision : Here , the value of the test statistic does not lies in the rejection region.

Therefore , fail to reject Ho.

Conclusion : There is not sufficient evidence to support the manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor.