Question

5. Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of an MP3 player were happy with their purchase. The advocacy group took a random sample of 1000 consumers who recently purchased this MP3 player and found that 400 were happy with their purchase. (a) Find a 95% lower bound for p (b) How large a sample size n should they take to estimate p with a 0.04 width and 90% confidence? (c) The advocacy group wants to determine whether the proportion p of consumers that are happy with their purchase is actually less than 0.5. Perform a hypothesis testing and state your conclusion using a 0.05 (d) Calculate the Type ll error probability at po 0.45

Answer 1

Proportion of customers happy with purchase, p = 400/1000 =0.4

a.)
Since sample is big, n = 1000, we will use z statistic to find lower bound of p

At 95% confidence level, z = 1.96

Lower Limit = p - z *

Vp* (1 - p)/i

Lower Limit = 0.4 - 1.96 *

V(0.4 0.6)/1000

Lower Limit = 0.36964 = 0.37

b.)

At 90% Confidence, z = 1.64

Margin of Error, E = 0.04

n = Zip(1-p) le'

n = (1.64)² * 0.4 * 0.6 / 0.0016

n = 403.44 = 403

c)

Null Hypotheses, Ho: p >= 0.5

Alternate Hypotheses, Ha: p < 0.5

p- Po (1- Po 72 where p is the sample proportion Pois the hypothesized proportion nis the sample size

z = (0.4 - 0.5) /

V(0.5 0.5)/1000

z = -0.1 / 0.0158

z = -6.3291

For above value of z, p-value is very very small and it is less than 0.05.

Since p < 0.05, we can reject the Null Hypotheses.

Hence, we can say that proportion of consumers who are actually happy with their purchase is less than 0.5

d.)

Type 2 Error Occurs when Ho is actually False and we accept Ho

So, we need to find probability of Ho

Assuming $\alpha$ = 0.05 and given po = 0.45

z = (0.4 - 0.45) /

0.45 0.55)/1000

z = -3.178

P(Z < -3.178) = 0.000762 = 0.08

So P(Type 2 Error) = 0.08