Question

Previous studies suggest that the length of time that a FL resident has resided in the city is normally distributed with a population standard deviation of 11.95 months.

a) In order to construct a 94% confidence interval for the true mean time with a margin of error no greater than 2 months, how large of a sample should be gathered?

The critical value, z_c=  (round to two decimal places).

The sample size, n =  .

b) A random sample of 203 FL residents has a sample mean of time spent living in FL as 75 months.

Construct a 94% confidence interval for the true mean time residing in the city.

Confidence Interval is: (  ,  ).

Using the same data as above, test the claim that the mean length of FL residency is less than 90 months at the 1% significance level.

Which of the following are the correct null and alternative hypotheses for the question above?

Group of answer choices

Answer 1

a)
The following information is provided,
Significance Level, α = 0.06, Margin or Error, E = 2, σ = 11.95

The critical value for significance level, α = 0.06 is 1.88.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.88 * 11.95/2)^2
n = 126.18

Sample size = 126

b)
sample mean, xbar = 75
sample standard deviation, σ = 11.95
sample size, n = 203

Given CI level is 94%, hence α = 1 - 0.94 = 0.06
α/2 = 0.06/2 = 0.03, Zc = Z(α/2) = 1.88

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (75 - 1.88 * 11.95/sqrt(203) , 75 + 1.88 * 11.95/sqrt(203))
CI = (73.42 , 76.58)

c)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 90
Alternative Hypothesis, Ha: μ < 90

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (75 - 90)/(11.95/sqrt(203))
z = -17.88

P-value Approach
P-value = 0
As P-value < 0.01, reject the null hypothesis.