H0: Null Hypothesis:
= 2500
HA: Alternative Hypothesis:
> 2500
SE = s/
= 147/
= 21
Test statistic is:
t = (2542 - 2500)/21 = 2
= 0.01
ndf = 49 - 1 = 48
From Table, critical value of t = 2.4066
Since the calculated value of t = 2 is less than critical value of t = 2.4066, the difference is not significant. Fil to reject null hypothesis.
Conclusion:
The data do not support the claim that the frequency flier miles will increase average usage.
At the 10% level, do you have enough evidence for the socom 2. A credit card...
1) Use either the critical value or p-value method for testing hypotheses. 2) Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), and critical value(s). 3) State your final conclusion that addresses the original claim. Include a confidence interval as well and restate this in your original conclusion. A credit card company wondered whether giving frequent flier miles for every purchase would increase card usage. The population mean had been $2500 per year. A simple random...
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a grocery store owner wants to know if a new marketing
campaign will
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