f\left(n\right)=\left\{\frac{\frac{n}{2}\ \ \ \ if\ n\ is\ even}{3n+1\ \ \ \ \ \ if\ n\ is\ odd}\right\}
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Answer: 61,750 miles
Step-by-step explanation:
Given : The p-value of the tires to outlast the warranty = 0.96
The probability that corresponds to 0.96 from a Normal distribution table is 1.75.
Mean :
Standard deviation :
The formula for z-score is given by : -
Hence, the tread life of tire should be 61,750 miles if they want 96% of the tires to outlast the warranty.
Answer:
99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].
Step-by-step explanation:
We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.
A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.
The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;
P.Q. = ~
where, = sample mean for mail-order sales = $74.50
= sample mean for internet sales = $84.40
= sample standard deviation for mail-order purchases = $17.25
= sample standard deviation for internet purchases = $21.25
= sample of sales receipts for mail-order purchases = 16
= sample of sales receipts for internet purchases = 9
Also, =
= 18.74
The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ().
Now, 99% confidence interval for () is given by;
=
Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.
=
= [$-31.82 , $12.02]
Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].