Question

1. Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) gives a good barometer of the overall stock market. On January 31, 2006, 9 of the 30 stocks making up the DJIA increased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financial analyst claims we can assume that 30% of the stocks traded on the New York Stock Exchange (NYSE) went up the same day.

A sample of 65 stocks traded on the NYSE that day showed that 28 went up.

You are conducting a study to see if the proportion of stocks that went up is is significantly more than 0.3. You use a significance level of α=0.01α=0.01.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is is more than 0.3.
• There is not sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is is more than 0.3.
• The sample data support the claim that the proportion of stocks that went up is is more than 0.3.
• There is not sufficient sample evidence to support the claim that the proportion of stocks that went up is is more than 0.3.

2. You are concerned that nausea may be a side effect of Tamiflu, but you cannot just give Tamiflu to patients with the flu and say that nausea is a side effect if people become nauseous. This is because nausea is common for people who have the flu. From past studies you know that about 30% of people who get the flu experience nausea. You collected data on 1685 patients who were taking Tamiflu to relieve symtoms of the flu, and found that 553 experienced nausea. Use a 0.01 significance level to test the claim that the percentage of people who take Tamiflu for the relief of flu symtoms and experience nausea is greater than 30%.  

a) Identify the null and alternative hypotheses?

H0 Select an answer p = p ≠ p < p > p ≤ p ≥ μ = μ ≠ μ < μ > μ ≤ μ ≥  

H1:  Select an answer p = p ≠ p < p > p ≤ p ≥ μ = μ ≠ μ < μ > μ ≤ μ ≥  

b) What type of hypothesis test should you conduct (left-, right-, or two-tailed)?

• left-tailed
• right-tailed
• two-tailed

c) Identify the appropriate significance level.


d) Calculate your test statistic. Write the result below, and be sure to round your final answer to two decimal places.


e) Calculate your p-value. Write the result below, and be sure to round your final answer to four decimal places.


f) Do you reject the null hypothesis?

• We reject the null hypothesis, since the p-value is less than the significance level.
• We reject the null hypothesis, since the p-value is not less than the significance level.
• We fail to reject the null hypothesis, since the p-value is less than the significance level.
• We fail to reject the null hypothesis, since the p-value is not less than the significance level.

g) Select the statement below that best represents the conclusion that can be made.

• There is sufficient evidence to warrant rejection of the claim that the percentage of people who experience nausea is greater than 30%.
• There is not sufficient evidence to warrant rejection of the claim that the percentage of people who experience nausea is greater than 30%.
• The sample data support the claim that the percentage of people who experience nausea is greater than 30%
• There is not sufficient sample evidence to support the claim that the percentage of people who experience nausea is greater than 30%

h) Can we conclude that nausea a side effect of Tamiflu?

• Yes
• No

Answer 1

1) H₀: P = 0.3

H₁: P > 0.3

= 28/65 = 0.431

The test statistic z = ( - p)/sqrt(p(1 - p)/n)

= (0.431 - 0.3)/sqrt(0.3 * 0.7/65)

= 2.3

P-value = P(Z > 2.3)

= 1 - P(Z < 2.3)

= 1 - 0.9893 = 0.0107

Since the p-value is greater than the significance level (0.0107 > 0.01), so we should not reject the null hypothesis.

There is sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is more than 0.3.

2) H₀: p = 0.3

H₁: p > 0.3

B) right-tailed

C) The significance level = 0.01

D) = 553/1685 = 0.328

The test statistic z = ( - p)/sqrt(p(1 - p)/n)

= (0.328 - 0.3)/sqrt(0.3 * 0.7/1685)

= 2.51

E) P-value = P(Z > 2.51)

= 1 - P(Z < 2.51)

= 1 - 0.9940 = 0.006

F) Since the p-value is less than the significance level (0.006 < 0.01), so we should reject the null hypothesis.

We reject the null hypothesis since the p-value is less than the significance level.

G) There is not sufficient evidence to warrant rejection of the claim that the percentage of people who experience nausea is greater than 30%.

H) Yes.