Answer:
Step-by-step explanation:
a) 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 = 9
For the alternative hypothesis,
µ ≠ 9
This is a 2 tailed test
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 9
x = 9.9
σ = 2.5
n = 50
z = (9.9 - 9)/(2.5/√50) = 2.55
Looking at the normal distribution table, the probability corresponding to the area above the z score is 1 - 0.99461 = 0.00539
Recall, population mean is 9
The difference between sample sample mean and population mean is 9.9 - 9 = 0.9
Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 9 - 0.9 = 8.1
the x value for the right tail is 9 + 0.9 = 9.9
These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area above the z score as z = 0.00539
We would double this area to include the area in the left tail of z = - 2.55. Thus
p = 0.00539 × 2 = 0.01078
Since alpha, 0.01 < 0.01078, we would reject the null hypothesis
But we are told to use the critical value approach, then
Since α = 0.01, the critical value is determined from the normal distribution table.
For the left, α/2 = 0.01/2 = 0.005
The z score for an area to the left of 0.005 is - 2.575
For the right, α/2 = 1 - 0.005 = 0.995
The z score for an area to the right of 0.995 is 2.575
In order to reject the null hypothesis, the test statistic must be smaller than - 2.575 or greater than 2.575
Since - 2.55 > - 2.575 and 2.55 < 2.575, we would reject the null hypothesis. This corresponds to our previous decision.
b) we would assume normal distribution because the sample size is sufficiently large and the population standard deviation is known.
c) Confidence interval is written in the form,
(Sample mean ± margin of error)
Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.
Margin of error = z × σ/√n
The z score for confidence level of 99% is 2.58
Margin of error = 2.58 × 2.5/√50 = 0.91
Confidence interval = 9.9 ± 0.91
The lower end of the confidence interval is
9.9 - 0.91 = 8.99
Therefore, p = 9 will be contained in the interval.
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