Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 106
For the alternative hypothesis,
µ > 106
It is a right tailed test because of >.
Since no population standard deviation is given, the distribution is a student's t.
Since n = 241,
Degrees of freedom, df = n - 1 = 241 - 1 = 240
t = (x - µ)/(s/√n)
Where
x = sample mean = 109
µ = population mean = 106
s = samples standard deviation = 8.3
a) test statistic, t = (109 - 106)/(8.3/√241) = 5.61
We would determine the p value using the t test calculator. It becomes
p = 0.00001
Since alpha, 0.05 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 5% level of significance, this improvement seem to be practicallysignificant.
b) Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
s = sample standard deviation
n = number of samples
From the information given, the population standard deviation is unknown, hence, we would use the t distribution to find the z score
df = 240
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
z = 1.6512
Margin of error = 1.6512 × 8.3/√241
= 0.88
Confidence interval = 109 ± 0.88
