Question

2.5 A Spar retailer observed a random sample of 161 customers and found that 69 customers paid for their grocery purchases by cash and the remainder by credit card. Question: Construct a 90% confidence interval for the actual percentage of customers who pay cash for their grocery purchases.

5 points

a) The actual percentage of customers who pay cash for their grocery purchases lies between 42.81% and 43.90%.

b) The actual percentage of customers who pay cash for their grocery purchases lies between 36.44% and 49.28%.

c) The actual percentage of customers who pay cash for their grocery purchases lies between 42.12% and 41.80%.

d) The actual percentage of customers who pay cash for their grocery purchases lies between 37.85% and 47.87%.

2.6 A survey amongst a random sample of 250 male and female respondents was conducted into their music listening preferences. Each respondent was asked whether they enjoy listening to jazz. Of the 140 males surveyed, 46 answered ‘Yes’. Of the 110 female respondents, 21 answered ‘Yes’. Is there statistical evidence at the 5% level of significance that males and females equally enjoy listening to jazz? Question: Formulate the Null and Alternative Hypothesis for this problem and tick the correct answer below.

2 points

a) H0: μ1 - μ2 ≠ 0 vs H1: μ1 - μ2 > 0

b) H0: μ1 - μ2 = 0 vs H1: μ1 - μ2 ≠ 0

c) H0: μ1 - μ2 < 0 vs H1: μ1 - μ2 < 0

d) H0: μ1 - μ2 ≤ 0 vs H1: μ1 - μ2 < 0

2.7 Make use of the information provided in the previous question and calculate the z-statistic using Z-test and tick the correct answer below.

5 points

a) z-stat = 1.96

b) z-stat = 2.19

c) z-stat = 2.44

d) z-stat = -2.01

2.8 Based on your empirical evidence in the previous question make a statistical conclusion whether there is statistical evidence at the 5% level of significance that males and females equally enjoy listening to jazz. Tick the correct answer below.

5 points

a) None of these answers is correct.

b) Reject the Null hypothesis. The alternative is probably true that the male and female proportions differ.

c) Fail to reject the Null hypothesis, the Null is probably true that the male and female proportions is the same.

d) Accept the Null hypothesis because the statistic is very close to the z-critical = 1.96.

Answer 1

2.5)

Given
p̂ = 0.4286           ....... Sample Proportion
n = 161           ....... Sample Size

For 90% Confidence interval

α = 0.1,      α/2 = 0.05
From z tables of Excel function NORM.S.INV(α/2) we find the z value
z = NORM.S.INV(0.05) = 1.645
We take the positive value of z

Confidence interval is given by


= (0.3644, 0.4928)

= (36.44%, 49.28%)

Answer ;

b) The actual percentage of customers who pay cash for their grocery purchases lies between 36.44% and 49.28%.

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2.6)

	n	Proportion (p)
Males	n1 = 140	p1 = 0.3286
Females	n2 = 110	p2 = 0.1909

The null and alternative hypotheses are
Ho :  P1 - P2 = 0
Ha :  P1 - P2 ≠ 0
where P1, P2 are the population proportions for males and females who enjoy listening to jazz respectively

Answer :  

Note to student : This is a hypotheses test for two proportions. 'p' is the symbol for proportion. But the choices have which is a symbol for mean. Yet giving the answer as 'b' as it is the closest correct option.(if we ignore the symbol)

b) H0: μ1 - μ2 = 0 vs H1: μ1 - μ2 ≠ 0

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2.7)

Calculations for z-statistic

Pooled Proportion

p̂ = 0.268

Standard Error (SE)

SE = 0.0564

z-statistic

z-statistic = z = 2.44

Answer:

c)   z-stat = 2.44

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2.8)

α = 0.05

P-value
For z = 2.44 we find the Equal to p-value using z tables or Excel function NORM.S.DIST
p-value = 2*NORM.S.DIST(-abs(2.44), TRUE)
p-value = 0.0147

Decision
0.0147 < 0.05
that is p-value is less than alpha.
Hence we Reject Ho.

Answer:
b) Reject the Null hypothesis. The alternative is probably true that the male and female proportions differ.

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