pank found that in reCe standard deviation of $183.34. Do these data provide statistical evidence that the average monthly charge has increased? Formulate the appropriate hypothesi there sufficient evidence at the 0.05 level of significance that the average monthly charge has increased? etermine the null hypothesis, Ho, and the alternative hypothesis, H, Question Viewer Type whole numbers)

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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A bank found that in recent years, the average monthly charge on its credit card was $1,400. With an improving economy, they suspect that this amount has increased. A sample of 40 customers resulted in an average monthly charge of $1,425.58 with
a standard deviation of $183.34. Do these data provide statistical evidence that the average monthly charge has increased? Formulate the appropriate hypothesis test and draw a conclusion. Use a level of significance of 0.05.
Is there sufficient evidence at the 0.05 level of significance that the average monthly charge has increased?
Determine the null hypothesis, Ho, and the alternative hypothesis, H1.
Ho
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(Type whole numbers.)
Transcribed Image Text:A bank found that in recent years, the average monthly charge on its credit card was $1,400. With an improving economy, they suspect that this amount has increased. A sample of 40 customers resulted in an average monthly charge of $1,425.58 with a standard deviation of $183.34. Do these data provide statistical evidence that the average monthly charge has increased? Formulate the appropriate hypothesis test and draw a conclusion. Use a level of significance of 0.05. Is there sufficient evidence at the 0.05 level of significance that the average monthly charge has increased? Determine the null hypothesis, Ho, and the alternative hypothesis, H1. Ho Question Viewer (Type whole numbers.)
Expert Solution
Step 1

Here,

The Null hypothesis is H0μ=1400

The Alternate hypothesis is H1:μ>1400

Here,

mean(x¯) = 1425.58

Standard deviation(δ)=183.34

total number of observation (n) = 40

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