Question

ATMs must be stocked with cash to satisfy customers making withdrawals over the weekend. Suppose that at a particular branch the population mean amount of money withdrawn from ATMs per customer transaction over the weekend is $160 with a population standard deviation of $30. 1) If a random sample of 36 customer transactions indicates that the sample mean withdrawal amount is $172, a) Is there evidence to believe that the population mean is not equal to $160? (  =0.05). b) Compute the test p-value. c) What is the statistical decision if you use a 0.01 level of significance? 2) If a random sample of 36 customer transactions indicates that the sample mean withdrawal amount is $172 a) Is there evidence that the population mean is higher than 160 dollars? Use  = 0.05. b) What is the p-value?

Answer 1

Solution :

1)

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 160

H_a :    160

= 172

= 160

= 30

n = 36

Test statistic = z

= ( - ) / / n

= (172 - 160) / 30 / 36

= 2.4

Test statistic = 2.4

P(z > 2.4) = 1 - P(z < 2.4) = 1 - 0.9918 = 0.0082

P-value = 2 * 0.0082 = 0.0164

= 0.01  

P-value >

Fail to reject the null hypothesis .

2)

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 160

H_a : > 160

P(z > 2.4) = 1 - P(z < 2.4) = 1 - 0.9918 = 0.0082

P-value = 0.0082

= 0.01  

P-value <

Reject the null hypothesis .

There is sufficient evidence that the population mean is higher than 160 dollars .