Jan notes that 6 out of every 18 cars that pass her on the
1 answer:
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<h2><u>Answer with explanation</u>:</h2>
Let be the distance traveled by deluxe tire .
As per given , we have
Null hypothesis :
Alternative hypothesis :
Since is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.
Test statistic :
where, n= sample size
= sample mean
= Population mean
=sample standard deviation
For , we have
By using z-value table,
P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23) [∵P(Z≤-z)=1-P(Z≤z)]
=1-0.9871=0.0129
Decision : Since p value (0.0129) < significance level (0.05), so we reject the null hypothesis .
[We reject the null hypothesis when p-value is less than the significance level .]
Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.