Question

Confidence Interval & Hypothesis Testing for a Single Population

Dominick Aldo owns and operates Carolina’s which is an Italian restaurant in New York. The file Single Population.xlsx is shown below and contains the amount of time that 90 tables were seated.

1. Develop a 90% confidence interval for this random sample and interpret the results.
2. Test the hypothesis that the average table time exceeds 98 minutes using α=.05.
3. What is the p-value? Interpret the results.

Be sure that your project shows the following steps

1. Null and alternative hypothesis
2. Determine which distribution to use for the test statistic.
3. Using data provided, calculate necessary sample statistics.
4. Draw a conclusion and interpret the decision.

Times for a Table

70
80
100
90
75
60
70
75
110
90
100
75
60
75
80
75
165
60
75
90
110
95
110
150
50
70
90
110
165
60
145
80
115
75
50
90
110
90
100
110
110
85
70
145
120
130
80
90
105
105
100
95
80
100
120
130
75
90
70
125
80
90
95
120
150
195
70
80
110
80
80
85
90
150
60
90
135
170
85
90
120
105
60
70
50
80
100
90
135
120

Answer 1

Calculate the sanple mean and sample standard deviation we get,

Sample size : n = 90

Mean : =  96.78

Standard deviation : s = 29.11

A)

c = 90%

formula for confidence interval is

Where tc is the t critical value for c=90 % with df=n-1 = 90-1 = 89

using t table we get critical value as

tc = 1.662

91.68 <    < 101.88

We get confidence interval as ( 91.68 , 101.88 )

B)

Answer:

n= 90 ,  = 98

= 96.78 , s = 29.11

= 0.05

null and alternative hypothesis is

Ho:   98

Ha:   > 98

here sample size is 90 which is greater than 30 hence, according to central limit theorem the distribution is approximately normal.

here population standard deviation is unkonown hence we are using t distribution.

formula for test statistics is

t = -0.39759

test statistics : t = -0.398

now calculate P-Value for this one tailed test with df= n-1 = 90 - 1 = 89, using following

excel command we get p-value as,

P-Value = 0.6541

decision rule is,

Reject Ho if (P-Value) < ( )

Here, ( P-Value = 0.6541 ) > ( = 0.05 )

Hence,

Fail to reject the null hypothesis

Therefore there is not enough evidence to support the claim that the average table time exceeds 98 minutes