Question

Los records pasados de un supermercado demuestran que sus clientes gastan un promedio de $65 por visita. Recientemente la gerencia inició una campaña promocional en la que cada cliente recibe puntos basado en el total de su compra y estos puntos se pueden usar para comprar artículos en la tienda. La gerencia espera, como resultado de esta campaña, que los clientes compren más. Para verificar si la campaña funcionó, el gerente toma una muestra de 12 clientes. Los siguientes son los totales de sus compras:

$88      69    141    35    106    45    32    51    78    54    110    83

         Asuma que la distribución del total de las compras es aproximadamente Normal. Utilizando     un nivel de significancia de 5%, ¿cree usted que la campaña promocional funcionó?

The past records of a supermarket demonstrated that their clients spend a mean of $65 per visit. Recently the manager started a promotional campaign in which each client receives points based on the total of their purchases and these points can be used to buy products in the store. The manager expects, as a result of this campaign, for clients to spend more. To verify that the campaign was successful, the manager takes a sample of 12 clients. The following are the total of their purchases:

$88      69    141    35    106    45    32    51    78    54    110    83

Assume that the distribution of the total of the purchases is approximately Normal. Using a level of significance of 5%, do you believe the promotion campaign worked?

Answer 1

The data is provided as:

	Data
	88
	69
	141
	35
	106
	45
	32
	51
	78
	54
	110
	83
Count	12
Mean	74.3333
SD	33.2766

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 65

Ha: μ > 65

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.796.

The rejection region for this right-tailed test is

(3) Test Statistics

The t-statistic is computed as follows:

(4) The decision about the null hypothesis

Since it is observed that t=0.972≤tc​=1.796, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1761, and since p=0.1761≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is greater than 65, at the 0.05 significance level. Hence the promotion campaign was not successful.

Graphically

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