Question 1 Mean = 114.8 mm Hg Standard Deviation = 13.1 mm Hg

Question 1
Mean = 114.8 mm Hg
Standard Deviation = 13.1 mm Hg
Sample size = 4
X = 140 mm Hg
Z = (140-114.8)/(13.1/40.5)= 3.847
P (Z<3.847) = 0.99994
Requisite probability = (1-0.99994) =0.0001 (Option D)
Question 2
Standard error = Standard deviation /Sample size 0.5 = 25/(300.5)= 4.56 (Option C)
Question 3
Considering that population standard deviation is not known, hence the t critical value needs to
be determined with 95% confidence and degrees of freedom as (30-1) =29. Further, two tail
value would be used for confidence interval. This value comes out as 2.045 (Option E).
Question 4
Margin of error = 4.56*2,045 = 9.33 (Option D)
Question 5
Confidence Interval = 140+/- 9.33 = (131.149)
Hence, both c and d are correct (Option F)
Question 6
The claim to be tested is shown as alternative hypothesis. Also, population mean is denoted by μ.

Hence, the correct null hypothesis is Option D
Question 7
The claim to be tested is shown as alternative hypothesis. It needs to be tested if population mean
is less than 0.16 or not.
Hence, the correct alternative hypothesis is Option B
Question 8
Since the population standard deviation is unknown, hence t statistic would be used. Also, the
test is left tailed. Degrees of freedom = 9-1 = 8
The requisite critical value is option B (-1.86)
Question 9
T =(0.16-0.0793)/(0.0048/9)0.5 = -3.49 (Option C)
Question 10
Since the computed value of t statistic exceeds the t critical value in magnitude, hence the null
hypothesis would be rejected and alternative hypothesis would be accepted. The claim is true
and hence Option A
Question 11
The test statistic would be altered to z value in the given case as population standard deviation is
known. As a result, left tail critical value for 95% confidence is -1.65 (Option A).