Customers spending on groceries follows a normal distribution with the variance of $3.50. The amount customers spend on non-groceries in same store follows a normal distribution with the variance of $4.10. The store offers a 15% savings on groceries and 20% savings on non-groceries. Assume that expenditure on groceries and non-groceries are independent of each other.
Out of a random sample of 12 customers, the sample average saving is $9.63. Test the hypothesis that the population average saving exceeds $9.33.
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What I think here is I need to make use of the fact that both random variables are independent. So then, Variance [X+Y] =Var[X] + Var[Y] = $7.60 & hence standard error is Var/sqrt(n) = 7.6/sqrt(12) = 2.194.
To get my test statistic I am stating for the null Ho: sample mean - population mean > 0 or ($9.63-$9.33)/(s.e). I get a Z-score test statistic of 0.136 which would mean I reject at not only a 5% level but even higher.
I am not confident at all about this and feel I've done something weird or crazy wrong. The bit about the % of discount maybe relevant but not sure how to incorporate that either. Thankful for any help.
Out of a random sample of 12 customers, the sample average saving is $9.63. Test the hypothesis that the population average saving exceeds $9.33.
________________________________________________________________________________________________________________________________________________________
What I think here is I need to make use of the fact that both random variables are independent. So then, Variance [X+Y] =Var[X] + Var[Y] = $7.60 & hence standard error is Var/sqrt(n) = 7.6/sqrt(12) = 2.194.
To get my test statistic I am stating for the null Ho: sample mean - population mean > 0 or ($9.63-$9.33)/(s.e). I get a Z-score test statistic of 0.136 which would mean I reject at not only a 5% level but even higher.
I am not confident at all about this and feel I've done something weird or crazy wrong. The bit about the % of discount maybe relevant but not sure how to incorporate that either. Thankful for any help.