1. In the last few years, many research studies have shown that the purported benefits of...

1. In the last few years, many research studies have shown that the purported benefits of hormone replacement therapy (HRT) do not exist, and in fact, that hormone replacement therapy actually increases the risk of several serious diseases. A four-year experiment involving 4400 women was conducted at 38 medical centres. Half of the women took placebos and half took a prescription drug, a widely prescribed type of hormone replacement therapy. There were x₁ = 49 cases of dementia in the hormone group and x₂ = 26 in the placebo group. Is there sufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug? Test at the 1% level of significance. (Round your answers to two decimal places.)

Test statistics=

Rejection region= z >

2. Independent random samples of n₁ = 200 and n₂ = 200 observations were randomly selected from binomial populations 1 and 2, respectively. Sample 1 had 116successes, and sample 2 had 122 successes.

Calculate the standard error of the difference in the two sample proportions, (p̂₁ − p̂₂). Make sure to use the pooled estimate for the common value of p. (Round your answer to four decimal places.)

p-value=

z < =

z > =

Solutions

Expert Solution

#1.
p1cap = X1/N1 = 49/2200 = 0.0223
p1cap = X2/N2 = 26/2200 = 0.0118
pcap = (X1 + X2)/(N1 + N2) = (49+26)/(2200+2200) = 0.017

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

Rejection Region
This is right tailed test, for α = 0.01
Critical value of z is 2.33.
Hence reject H0 if z > 2.33

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.0223-0.0118)/sqrt(0.017*(1-0.017)*(1/2200 + 1/2200))
z = 2.69

#2.
p1cap = X1/N1 = 116/200 = 0.58
p1cap = X2/N2 = 122/200 = 0.61

pooled estimate, pcap = (X1 + X2)/(N1 + N2) = (116+122)/(200+200) = 0.5950

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.58-0.61)/sqrt(0.595*(1-0.595)*(1/200 + 1/200))
z = -0.61

P-value Approach
P-value = 0.5419
As P-value >= 0.05, fail to reject null hypothesis.

orchestra answered 1 year ago

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