[7 points] A sample of 78 anchor bolts was randomly selected and the shear strength (in kips) of each was measured, resulting in a sample mean of 4.25 and a sample standard deviation of 1.30. (a) Calculate a 98% confidence interval to estimate the true mean shear strength. [4 points] (b) Suppose that the shear strength for all bolts in the population was not normally distributed. Would the confidence interval found in part (a) still be valid

Step-by-step explanation:

We want to determine a 98% confidence interval to estimate the true mean shear strength of anchor bolts.

Number of sample, n = 78

Mean, u = 4.25

Standard deviation, s = 1.3

For a confidence level of 98%, the corresponding z value is 2.33.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

4.25 ± 2.33 × 1.3/√78

= 4.25 ± 2.33 × 0.147

= 4.25 ± 0.34

The lower end of the confidence interval is 4.25 - 0.34 = 3.91

The upper end of the confidence interval is 4.25 + 0.34 = 4.59

b) the confidence interval would still be valid because the number of samples is large(> 30)

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.