Question

. Insurance companies are interested in knowing the mean weight of cars currently licensed in the United States; they believe it is 3000 pounds. To see if the estimate is correct, they check a random sample of 80 cars. For that group, the mean weight was 2910 pounds with a standard deviation of 532 pounds. Is this strong evidence that the mean weight of all cars is not 3000 pounds? a. What is the population of interest? b. Describe in words. d. Perform the test using a significance level of 0.05 ( = 0.05) ii. P-value Due to the different methods, I rounded out to 2 decimal places. Select the best possible answer. iii. Conclusion in context h. Find a 90% confidence interval for . Due to the different methods, select the best possible answer. k. What sample size would allow us to increase our confidence level to 95% while reducing the margin of error to only 50 pounds?

Answer 1

Answer)

As the population s.d is unknown we will use t distribution to solve this problem

Null hypothesis Ho : u = 3000

Alternate hypothesis Ha : u not equal to 3000

A)

Population of interest is weight of cars currently licensed in the United States.

B)

Mean weight of cars currently licensed in the United States.

C)

Test statistics t = (sample mean - claimed mean)/(s.d/√n)

t = (2910 - 3000)/(532/√80) = -1.51

Degrees of freedom is = n-1 = 79

For 79 dof and -1.51 test statistics

P-value from t distribution is = 0.135

As 0.135 is > 0.05

We fail to reject the null hypothesis Ho

So we do not have enough evidence to conclude that mean is not equal to 3000

90% confidence interval

Critical value t from t table for 90% confidence level and 79 dof is 1.66

Error = t*s.d/√n = 1.66*532/√80 = 99

Confidence interval is given by

(Mean - moe, mean + moe)

[2811, 3009].

K)

We can use z table to estimate the sample size

Critical value z from z table for 95% confidence level is 1.96

Error = 50

50 = 1.96*532/√n

N = 435