Question

student	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Test score	67	67	87	89	87	77	73	74	68	72	58	98	98	70	77

Above we have the final averages of the last stats and I want to know if the class average falls within the boundaries of all my statistics classes for the past 20 years.

Find the sample size, mean, and standard deviation of the data above (Table 1). Then use the 8-step method for determining whether or not last year’s class meets or exceeds the historical standards for statistics.

	SIZE	MEAN	STANDARD DEVIATION
SAMPLE (last class)	15
POPULATION (all classes)	1,000	75	5.9

Answer 1

Solution:-

Data
Mean	77.46667
Standard Error	3.057959
Median	74
Mode	67
Standard Deviation	11.84342
Sample Variance	140.2667
Kurtosis	-0.58604
Skewness	0.463985
Range	40
Minimum	58
Maximum	98
Sum	1162
Count	15

SIZE	Size	Mean
SAMPLE (last class)	15	77.4667	11.8434
POPULATION (all classes)	1,000	75	5.9

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 75.0
Alternative hypothesis: u > 75.0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 3.058
DF = n - 1

D.F = 14
t = (x - u) / SE

t = 0.807

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 0.807.

Thus the P-value in this analysis is 0.217.

Interpret results. Since the P-value (0.217) is greater than the significance level (0.05), we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that last year’s class exceeds the historical mean.

Test for standard deviation

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis H₀: σ < 5.9

Alternative hypothesis H_A: σ > 5.9

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data. Using sample data, the degrees of freedom (DF), and the test statistic (X²).

DF = n - 1 = 15 -1

D.F = 14

We use the Chi-Square Distribution Calculator to find P(Χ² > 56.41) = 0.00

Interpret results. Since the P-value (0.00) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that last year’s class exceeds the historical standard deviation.