Answer:
(A) What is the z- score of the sample mean?
The z- score of the sample mean is 0.0959
(B) Is this sample significantly different from the population?
No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.
Step -by- step explanation:
(A) To find the z- score of the sample mean,
X = 75 which is the raw score
¶ = 74 which is the population mean
S. D. = 10.43 which is the population standard deviation of/from the mean
Z = [X-¶] ÷ S. D.
Z = [75-74] ÷ 10.43 = 0.0959
Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].
(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.
The estimated standard error of the mean is s/√n
S. E. = 16/√80 = 16/8.94 = 1.789
The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559
Setting up the hypotheses,
Null hypothesis: Sample is not significantly different from population
Alternative hypothesis: Sample is significantly different from population
Having gotten T- cal, T- tab is found thus:
The Critical (Table) t value is found using
- a specific alpha or confidence level
- (n - 1) degrees of freedom; where n is the total number of observations or items in the population
- the standard t- distribution table
Alpha level = 0.05
1 - (0.05 ÷ 2) = 0.975
Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;
The critical t value is 1.990
Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.
Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.