Answer:
Step-by-step explanation:
Hello!
To study if the average dollars spend per shopper is different than the expected population average of $84.5 there was a sample of 120 shoppers taken and their spending habit registered.
Then the study variable is:
X: Amount of money spent by a shopper ($)
This population standard deviation is σ= $16
Using the data information I've run a normality test, with p-value: 0.3056 against the test significance level α: 0.05, the decision is to not reject the null hypothesis so there is enough statistical evidence to conclude that the study variable has a normal distribution.
X~N(μ;σ²)
I've also calculated the sample mean and standard deviation:
X[bar]= $81.01
S= $19.62
a.
If the objective is to test that the true mean spending for the population is significantly different than the expected average of 84.5, the parameter of interest is the population mean μ and the hypothesis test will be two-tailed.
H₀: μ = 84.5
H₁: μ ≠ 84.5
α:0.05
The statistic is a standard normal for one sample:
![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~N%280%3B1%29)
p-value two tailed 0.016894
b.
The decision rule using the p-value method is:
If p-value ≤ α ⇒ Reject the null hypothesis.
If p-value > α ⇒ Do not reject the null hypothesis.
Since the p-value is less than the significance level, the decision is to reject the null hypothesis.
c.
Using a level of significance of 5% the null hypothesis was rejected. So there is enough statistical evidence to support the researcher's claim that the true mean spending for the population is significantly different than the expected average of $84.5.
I hope it helps!
