Question

A)

If n=15, ¯xx¯(x-bar)=37, and s=6, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

B).

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 394 drivers and find that 303 claim to always buckle up. Construct a 84% confidence interval for the population proportion that claim to always buckle up.

C)

A well-known brokerage firm executive claimed that 34 % of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that out of 133 randomly selected people, 49 of them said they are confident of meeting their goals.

Suppose you are have the following null and alternative hypotheses for a test you are running:

H0:p=0.34H0:p=0.34
Ha:p≠0.34Ha:p≠0.34

Calculate the test statistic, rounded to 3 decimal places

z=

Answer 1

A)

As the population s.d is unknown we will use t distribution to estimate the interval

Given info

n = 15

Mean = 37

S = 6

For n-1 = 14 dof and 80% confidence level, critical value t from t table is = 1.345

Margin of error (MOE) = t*s.d/√n = 1.345*6/√15= 2.08

Confidence interval is given by

(Mean - MOE, Mean + MOE)

[34.9 , 39.1].

We are 80% confident that the population mean (μ) falls between 34.9 and 39.1

B)

Given that,

Point estimate = sample proportion = = x / n = 303 / 394 = 0.769

1 - = 0.231

alpha = 1 - 0.84 = 0.16

Z_/2 = 1.405

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.405 * (((0.769* 0.231) / 394)

= 0.030

A 84% confidence interval for population proportion p is ,

- E < p < + E

0.769 - 0.030< p < 0.769 + 0.030

0.739 < p < 0.799

The 84% confidence interval for the population proportion p is : [0.739, 0.799]

C)

Point estimate = sample proportion = = x / n = 49/ 133 = 0.368

H0:p=0.34
Ha:p≠0.34

z=0.682