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 1919 phones from the manufacturer had a mean range of 12301230 feet with a standard deviation of 3131 feet. A sample of 1111 similar phones from its competitor had a mean range of 11901190 feet with a standard deviation of 4242 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=0.1α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 2 of 4 :  

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

step 3 t/|t| is </> ____

reject, fail to reject

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means by assuming equal population variances.

H₀: µ₁ = µ₂ versus H_a: µ₁ > µ₂

This is an upper tailed test.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 1230

X2bar = 1190

S1 = 31

S2 = 42

n1 = 19

n2 = 11

df = n1 + n2 – 2 = 19 + 11 – 2 = 28

α = 0.10

Critical value = 1.3125

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(19 – 1)* 31^2 + (11 – 1)* 42^2]/(19 + 11 – 2)

S_p² = 1247.7857

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (1230 – 1190) / sqrt[1247.7857*((1/19)+(1/11))]

t = 40/13.3831

t = 2.9888

t = 2.989

P-value = 0.0029

(by using t-table)

P-value < α = 0.10

So, we reject the null hypothesis

Conclusion:

There is sufficient evidence to conclude that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor.