question archive A car manufacturer, Swanson, claims that the mean lifetime of one of its car engines is greater than 220,008 miles, which is the mean lifetime of the engine of a competitor
A car manufacturer, Swanson, claims that the mean lifetime of one of its car engines is greater than 220,008 miles, which is the mean lifetime of the engine of a competitor
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A car manufacturer, Swanson, claims that the mean lifetime of one of its car engines is
greater than 220,008 miles, which is the mean lifetime of the engine of a competitor. The
mean lifetime for a random sample of 28 of the Swanson engines was x = 226,450 miles
with a standard deviation, s, of 11,504 miles. Test the Swanson's claim using a significance
level of α = 0.05.
A. P-value = 0.00314 < 0.05, reject the null hypothesis. There is enough evidence to conclude that the mean lifetime of this car engine is greater than 220,008 miles.
B. P-value = 0.00629 < 0.05, reject the null hypothesis. There is enough evidence to conclude that the mean lifetime of this car engine is greater than 220,008 miles.
C. P-value = 0.00414 < 0.05, reject the null hypothesis. There is enough evidence to conclude that the mean lifetime of this car engine is greater than 220,008 miles.
D. P-value = 0.00669 < 0.05, reject the null hypothesis. There is enough evidence to conclude that the mean lifetime of this car engine is greater than 220,008 miles.
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Required answer is option A
Step-by-step explanation
STEP 1: Null and Alternate hypothesis
?H0?:μ≤220008 vs H1?:μ>220008?
STEP 2: Test statistic t
Formula to calculate test statistic t is
?t=s/n?x?−μ? where x?=226450,μ=220008,n=28,s=11504 →11504/28?226450−220008? →2.963?
STEP 3: P-value
df = n-1 ?→28−1⇒27?
By t-distribution table
?⇒? p-value = 0.00314
STEP 4: Decision
Reject Ho since p-value (0.00314) < significance level 0.05.
STEP 5: Conclusion
There is enough evidence to conclude that the mean lifetime of this car engine is greater than 220,008 miles.
