Question

A blind taste test is conducted to determine which of two colas, Brand A or Brand B, individuals prefer. Individuals are randomly asked to drink one of the two types of cola first, followed by the other cola, and then asked to disclose the drink they prefer. Results of the taste test indicate that 59 of 100 individuals prefer Brand A. Complete parts a through c. (a) Conduct a hypothesis test (preferably using technology) HO. p-p0 versus HI : pf p0 for Po-o48, 0.49, 0.50, , 0.68, 0.69, 0.70 at the α significance. For which values of po do you not reject the null hypothesis? What do each of the values of Po represent? 0.05 level of Do not reject the null hypothesis for the values of Po btenand, inclusively. Type integers or decimals as needed.) (b) Construct a 95% confidence interval for the proportion of individuals who prefer Brand A. The lower bound is The upper bound is (Round to three decimal places as needed.) (c) Suppose you changed the level of significance in conducting the hypothesis test to α=0.01. what would happen to the range of values for p0 for which the null hypothesis is not rejected? Why does this make sense? Choose the correct answer below. O A. O B. The range of values would increase because the corresponding confidence interval would increase in size The range of values would decrease because the corresponding confidence interval would decrease in size O C. The range of values would decrease because the corresponding confidence interval would increase in size ( D. The range of values would increase because the corresponding confidence interval would decrease in size

Answer 1

Part a

Here, we have to conduct z test for population proportion.

H₀: p = p₀ versus H₁: p ≠ p₀

Where, p₀ = .48, .49,...., .70

Z = (P – p)/sqrt(pq/n)

Where, q = 1 – p

P is sample proportion.

We are given

x = 59

n = 100

P = x/n = 59/100 = 0.59

For p = .48, q = 1 – 0.48 = 0.52

Z = (0.59 – 0.48)/[0.48*0.52/100)

Z = 2.2018

P-value = 0.0277

(by using excel)

P-value < α = 0.05

So, we reject the null hypothesis

Let us see z tests for other values of p₀ by using excel.

p0	Z	P-value	Decision
0.48	2.2018	0.0277	Reject
0.49	2.0004	0.0455	Reject
0.50	1.8000	0.0719	Do not reject
0.51	1.6003	0.1095	Do not reject
0.52	1.4011	0.1612	Do not reject
0.53	1.2022	0.2293	Do not reject
0.54	1.0032	0.3158	Do not reject
0.55	0.8040	0.4214	Do not reject
0.56	0.6044	0.5456	Do not reject
0.57	0.4040	0.6862	Do not reject
0.58	0.2026	0.8394	Do not reject
0.59	0.0000	1.0000	Do not reject
0.60	-0.2041	0.8383	Do not reject
0.61	-0.4100	0.6818	Do not reject
0.62	-0.6181	0.5365	Do not reject
0.63	-0.8285	0.4074	Do not reject
0.64	-1.0417	0.2976	Do not reject
0.65	-1.2579	0.2084	Do not reject
0.66	-1.4777	0.1395	Do not reject
0.67	-1.7014	0.0889	Do not reject
0.68	-1.9294	0.0537	Do not reject
0.69	-2.1622	0.0306	Reject
0.70	-2.4004	0.0164	Reject

Do not reject the null hypothesis for the values of p₀ between 0.50 and 0.68, inclusively.

Part b

Confidence interval for Population Proportion

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 59

n = 100

P = x/n = 59/100 = 0.59

Confidence level = 95%

Z = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = .59 ± 1.96* sqrt(.59*(1 – .59)/100)

Confidence Interval = .59 ± 1.96* 0.0492

Confidence Interval = .59 ± 0.0964

Lower limit = .59 - 0.0964 = 0.494

Upper limit = .59 + 0.0964 = 0.686

Confidence interval = (0.494, 0.686)

Part c

Answer: A. The range of values would increase because the corresponding confidence interval would increase in size.

Explanation:

We know that the size or width of the confidence interval increases if we decrease the significance level. If size of confidence interval increases, the range of values would be increase.