Use Excel to find the p- value . (Round your answer to 4 decimal places.) p- val
ID: 3431861 • Letter: U
Question
Use Excel to find the p- value . (Round your answer to 4 decimal places.) p- value Calculate the value of Test statistic .(Carry out all intermediate calculations to at least four decimal places. Round your answer to 3 decimal places.) Zcalc The Scottsdale fire department aims to respond to fire calls in 5 minutes or less, on average. Response times are normally distributed with a standard deviation of 1.6 minute. Would a sample of 18 fire calls with a mean response time of 5 minutes 15 seconds provide sufficient evidence to show that the goal is not being met at a = 0.01? Use Excel to find the p- value . (Round your answer to 4 decimal places.) p- value The average weight of a package of rolled oats is supposed to be at least 23 ounces .A sample of 18 packages shows a mean of 22.8 ounces with a standard deviation of 0.49 ounce.Explanation / Answer
Answers: Problem1
P=0.0369
Problem2
Z= 0.663
P=0.2537
Problem 1
T test used
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
23
Level of Significance
0.05
Sample Size
18
Sample Mean
22.78
Sample Standard Deviation
0.49
Intermediate Calculations
Standard Error of the Mean
0.1155
Degrees of Freedom
17
t Test Statistic
-1.9049
Lower-Tail Test
Lower Critical Value
-1.7396
p-Value
0.0369
Reject the null hypothesis
Problem
Population sd known Z test used
Z Test of Hypothesis for the Mean
Data
Null Hypothesis m=
5
Level of Significance
0.01
Population Standard Deviation
1.6
Sample Size
18
Sample Mean
5.25
Intermediate Calculations
Standard Error of the Mean
0.3771
Z Test Statistic
0.6629
Upper-Tail Test
Upper Critical Value
2.3263
p-Value
0.2537
Do not reject the null hypothesis
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
23
Level of Significance
0.05
Sample Size
18
Sample Mean
22.78
Sample Standard Deviation
0.49
Intermediate Calculations
Standard Error of the Mean
0.1155
Degrees of Freedom
17
t Test Statistic
-1.9049
Lower-Tail Test
Lower Critical Value
-1.7396
p-Value
0.0369
Reject the null hypothesis
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