A.) use t-test to test at the 5% level of significance to reject/accept the null

Question

A.) use t-test to test at the 5% level of significance to reject/accept the null hypothesis
B.) do you accept or fail to reject the null hypothesis and who is the winner/loser? Paige and Frankie are competing in a friendly rivalry of who can make the most profit from selling lemonade. The table below gives the sample of quantity of sales of glasses of lemonade each made acrOSS 5 days (suppose sales are normally distributed). Paige sells her lemonade at 50 cents each and Frankie sells hers at 70 cents each Day Paige Frankie 2210 150 3 90 90 4150100 5 240 150

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu1 - mu2 = 0 Vs H1 : mu1 - mu2 not= 0

where mu1 and mu2 are two population means of Paige and Frankie respectively.

Assume alpha = level of significance = 5% = 0.05

Here sample data is given also sample size is too small so we use two sample t test.

We can do two sample t-test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 2 Sample t --> Samples in differnt columns --> First : select Paige data --> Second : Select Frankie data --> Assume equal variances --> Options --> Confidence level : 95.0 --> Test mean : 0.0 --> Alternative : not equal --> ok --> ok

Two-Sample T-Test and CI: paige, frankie

Two-sample T for paige vs frankie

N Mean StDev SE Mean
paige 5 170.0 57.9 26
frankie 5 120.0 28.3 13

Difference = mu paige - mu frankie
Estimate for difference: 50.0
95% CI for difference: (-16.4, 116.4)
T-Test of difference = 0 (vs not =): T-Value = 1.74 P-Value = 0.121 DF = 8
Both use Pooled StDev = 45.6

Test statistic = 1.74

P-value = 0.121

Pvalue > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that the two population means are equal.

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