Question

Randomly selected students were given five seconds to estimate the value of a product of numbers with the results shown below.
Estimates from students given 1×2×3×4×5×6×7×8:

10000, 2040, 750, 4000, 42200, 6000, 1500, 5000, 500, 5000

Estimates from students given 8×7×6×5×4×3×2×1:

100000, 10000, 52836, 1200, 450, 100000, 200, 2050, 1500, 400

Use a 0.05 significance level to test the following claims:

Claim: the two populations have equal variances.

The test statistic is  
The larger critical value is  
The conclusion is
A. There is sufficient evidence to reject of the claim that the two populations have equal variances. (So, we can assume the variances are unequal.)
B. There is not sufficient evidence to reject the claim that the two populations have equal variances. (So, we can assume the variances are equal.)

Claim: the two populations have the same mean.

The test statistic is  
The positive critical value is  
The negative critical value is  
The conclusion is
A. There is not sufficient evidence to reject the claim that the two populations have the same mean.
B. There is sufficient evidence to reject the claim that the two populations have the same mean.

Answer 1

Solution:-

1)

Set 1		Set 2
Mean	7699	Mean	26863.6
Standard Error	3940.274	Standard Error	13200.76
Median	4500	Median	1775
Mode	5000	Mode	100000
Standard Deviation	12460.24	Standard Deviation	41744.47
Sample Variance	1.55E+08	Sample Variance	1.74E+09
Kurtosis	8.560508	Kurtosis	0.006698
Skewness	2.861455	Skewness	1.318767
Range	41700	Range	99800
Minimum	500	Minimum	200
Maximum	42200	Maximum	100000
Sum	76990	Sum	268636
Count	10	Count	10

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

Null hypothesis H₀: σ_A² = σ_B²

Alternative hypothesis H_A: σ_A² σ_B²

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 (F).

DF₁ = n₁ - 1 = 10 -1

D.F₁ = 9

DF₂ = n₂ - 1 = 10 -1

D.F₂ = 9

Test statistics:-

F = 0.09

F_Critical = + 0.315

Rejection region is F < 0.315

Since the first sample had the smaller standard deviation, this is a left-tailed test.

p value for the F distribution = 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.

A. There is sufficient evidence to reject of the claim that the two populations have equal variances. (So, we can assume the variances are unequal.)

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

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

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 13776.278
DF = 18
t = [ (x₁ - x₂) - d ] / SE

t = - 1.39

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 18 degrees of freedom is more extreme than -1.39; that is, less than -1.39 or greater than 1.39.

Thus, the P-value = 0.181.

Interpret results. Since the P-value (0.181) is greater than the significance level (0.05), hence we failed to reject null hypothesis.

A) There is not sufficient evidence to reject the claim that the two populations have the same mean.