Question

Every year,the Hometown Hardware Store expects an average family to spend at least $606.40 on springtime home repairs. A new analyst in the store disputes this and challenges that the average spending per family is less than this amount.

The analyst conducts a test on the basis of a random sample of 30 households, using a 10% significance level, resulting in a sample mean of $589.95. This is normally distributed using a population standard deviation of $65.

Critical Value = ??   2 decimal places

Statistical Value = ??    3 decimal places

pvalue = ??

Decision: = ??
Reject or Do Not Reject

Please Provide way, calculations, and sketch. And pleaseeee don't answer if you don't know!

Answer 1

Ho: u >= 606.4
h1: u < 606.4 (left tailed)

alpha = 10%
critical value, -z(a) = -z(0.1) = -1.28

test statistic, z = (mean-u)/(sd/sqrt(n))
z = (589.95-606.4)/(65/sqrt(30))
z = -1.386

p-value = P(Z<z)
p-value = P(z<-1.3862)
p-value = NORMSDIST(-1.3862)
p-value = 0.082842921

Decision = With p<0.10, i reject the null hypothesis at 10% level of significance
there is no sufficient evidence to support the claim that u >= 606.4.