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 1200 feet with a standard deviation of 43 feet. A sample of 16 similar phones from its competitor had a mean range of 1130 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=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

The [tex]z_\alpha[/tex] score against which we are going to compare is the z-score corresponding to a level of confidence 0.1, that is to say, the value z for which the area under the normal curve to the right of z is 0.1.

Either by looking up in a table or by using the computer, we find that this value is 1.282 .

The z-score for the test is given by

[tex]z=\frac{(\bar x_1-\bar x_2)-(\mu_1-\mu_2)}{\sqrt{(s_1^2/n_1)+(s_2^2/n_2)}}[/tex]

where

[tex]\bar x_1[/tex] is the mean of sample 1

[tex]s_1[/tex] is the standard deviation of sample 1

[tex]n_1[/tex] is the size of sample 1

[tex]\bar x_2[/tex] is the mean of sample 2

[tex]s_2[/tex] is the standard deviation of sample 2

[tex]n_2[/tex] is the size of sample 2

We then have

[tex]z=\frac{(1200-1130)-0}{\sqrt{(43^2/6)+(41^2/16)}}=2.8859[/tex]

Since the z-score is greater than 1.282 we can reject the null hypothesis.

So, the results do support the manufacturer's claim.