Question

A consultant is tasked with studying the puchasing habits of customers in an Indonesian coffee shop...

A consultant is tasked with studying the puchasing habits of customers in an Indonesian coffee shop which specialises in Kopi Luwak. They select a random sample of 60 customers during a certain week and find that the mean amount spent was 291,000 rupiah, with a standard deviation of 60,200 rupiah. They also found that 32 customers would 'strongly recommend' the coffee shop to their family and friends.

(a) At the 0.05 level of significance, is there evidence that the population mean amount spent was different from 261,000 rupiah? Show all your working and clearly indicate the assummptions used.

(b) At the 0.05 level of significance, is there evidence that more than 50% of all the customers say they would 'strongly recommend' the coffee shop to their family and friends? Show all your workings and assumptions used.

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Answer #1

a)

Sample size = n = 60

Sample mean = ar{x} = 291000

Standard deviation = s = 60200

Claim: The population mean amount spent was different from 261,000 rupiah.

The null and alternative hypothesis is

170 : μ-261000

H1 : μメ261000

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

t=rac{ar{x}-mu }{rac{s}{sqrt{n}}}

291000 - 261000 30000 = 3.86 = 7771.787

Degrees of freedom = n - 1 = 60 - 1 = 59

Critical value = 2.001 ( Using t table)

Test statistic > critical vaue we reject null hypothesis.

Conclusion: The population mean amount spent was different from 261,000 rupiah.

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b)

Claim: More than 50% of all the customers say they would 'strongly recommend' the coffee shop to their family and friends.

The null and alternative hypothesis is

H0: P leq 0.50

H1: P > 0.50

Level of significance = 0.05

Sample size = n = 60

x = 32

32 n 60 -_ = 0.53

Test statistic is

z=rac{hat{p}-P_{0}}{sqrt{P_{0}*(1-P_{0})/n}}

z=rac{0.53-0.50}{sqrt{0.50*(1-0.50)/60}}=rac{0.03333}{sqrt{0.004167}}=0.52

P-value = P(Z >0.52) = 0.3028

P-value > 0.05 we fail to reject null hypothesis.

Conclusion: NOT more than 50% of all the customers say they would 'strongly recommend' the coffee shop to their family and friends.

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