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 1111 phones from the manufacturer had a mean range of 13901390 feet with a standard deviation of 3333 feet. A sample of 2020 similar phones from its competitor had a mean range of 13601360 feet with a standard deviation of 3030 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=0.05α=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.
Step 2 of 4:
Compute the value of the t test statistic. Round your answer to three decimal places.
Step 3 of 4:
Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places.
Step 4 of 4:
State the test's conclusion
Step 1 of 4:
H0: Null Hypothsis: ( The calling range (in feet) of its 900-MHz cordless telephone of the manufacturer is not greater than that of its leading competitor)
HA: Alternative Hypothsis: ( The calling range (in feet) of its 900-MHz cordless telephone of the manufacturer is greater than that of its leading competitor.) (Claim)
Step 2 of 4:
n1 = 11
1 = 1390
s1= 33
n2 = 20
2 = 1360
s2 = 30
Pooled Standard Deviation is given by:
Test Statistic is given by:
Step 3 of 4:
= 0.05
df = 11 + 20 - 2 = 29
One Tail Test - Right Side
From Table, critical value of t = 1.699
Decision Rule:
Reject Null Hypothesis if t is greater than 1.699
Step 4 of 4:
Since calculated value of t = 2.572 is greater than critical value
of t = 1.699, the difference is significant. Reject null
hypothesis.
Conclusion:
The datasupport the claim that the calling range (in feet) of its
900-MHz cordless telephone of the manufacturer is greater than that
of its leading competitor.
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