Do sit down restaurant franchises and fast food franchises differ significantly
ID: 3361042 • Letter: D
Question
Do sit down restaurant franchises and fast food franchises differ significantly in stock price? Specifically, is the average stock price for sit-down restaurants less than the average stock price for fast food restaurants? If sit down restaurants are in group 1 and fast food restaurants are in group 2, the hypotheses for this scenario are as follows: Null Hypothesis: 1 2, Alternative Hypothesis: 1 < 2. In a random sample of 39 sit down restaurants, you find that the average stock price is $243.695 with a standard deviation of $15.8063. For 60 fast food restaurants, the average stock price is $251.592 with a standard deviation of $18.5862. Conduct a two independent sample t-test. What is the test statistic and p-value for this test? Assume the population standard deviations are the same. Question 19 options: 1) Test Statistic: -2.188, P-Value: 0.0155 2) Test Statistic: 2.188, P-Value: 0.9845 3) Test Statistic: 2.188, P-Value: 0.0155 4) Test Statistic: -2.188, P-Value: 0.9845 5) Test Statistic: -2.188, P-Value: 0.031
Explanation / Answer
Sit down restaurants:
n1 = 39
Mean= µ1 = $243.695
Standard deviation = s1 = $15.8063
Fast food :
n2 = 60
Mean= µ2 = $251.592
Standard deviation = s2 = $18.5862
H0 : µ1 = µ2
H1 : µ1 < µ2
Since the population variances are equal, the t- statistic is given by:
t = x1 - x2 / sp(1/n1 + 1/n2)
where sp is the pooled standard deviation.
Sp2 = (n1-1)s12 + (n2-1)s22 / (n1 + n2 -2)
=(38*249.83912 + 59*345.44683)/ 97
=307.9923
sp = 17.54971
t =-7.897/(17.54971*0.20568)
t = -2.18776
To obtain the p-value, use R software. Use the following code
pt(t,df=)
in this case, pt(-2.188,df=39+60-2)
p-value = 0.01553558
Since, p-value(0.01553) < 0.05, we reject the null hypothesis and conclude that the average stock price for sit-down restaurants less than the average stock price for fast food restaurants.
Answer: 1) t = -2.188 and p-value= 0.0155
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