A gum manufacturer advertises that 4 out of 5 dentists recommend sugarless gum.

Question

A gum manufacturer advertises that 4 out of 5 dentists recommend sugarless gum. A consumer advocacy group surveyed 100 dentists and found that 74 of them recommended sugarless gum.

A- Estimate a 99% confidence interval for the true proportion.

B- Conduct a hypothesis test to see if the proportion is consistent with the manufacturer’s claim, using a 99% confidence level.

C- If you want to estimate the difference to within 1.0%, what sample size should you use (99% confidence level)?

D- What is the observed significance level (p-value)?

Explanation / Answer

A- Estimate a 99% confidence interval for the true proportion.

Answer:

We are given

Sample size = n = 100

Number of successes = x = 74

Sample proportion = x/n = 74/100 = 0.74

Confidence level = 99%

Critical Z value = 2.5758

(By using z-table)

Confidence interval formula is given as below:

Confidence interval = P -/+ Z*sqrt[P(1 – P)/n]

Confidence interval = 0.74 -/+ 2.5758*sqrt[0.74(1 – 0.74) / 100]

Confidence interval = 0.74 -/+ 0.1130

Lower limit = 0.74 – 0.1130 = 0.6270

Upper limit = 0.74 + 0.1130 = 0.8530

Confidence interval = (0.6270, 0.8530)

B- Conduct a hypothesis test to see if the proportion is consistent with the manufacturer’s claim, using a 99% confidence level.

Answer:

Here, we have to use Z test for population proportion. The null and alternative hypotheses for this test are given as below:

H0: p = 0.80

Versus

Ha: p 0.80

We are given

Sample size = n = 100

Number of successes = x = 74

Sample proportion = P = x/n = 74/100 = 0.74

Confidence level = 99%, (c = 0.99), = 1 – 0.99 = 0.01

Critical Z value = -/+ 2.5758

(By using z-table)

Test statistic formula is given as below:

Z = (P – p) / sqrt(pq/n)

Z = (0.74 – 0.80) / sqrt(0.80*0.20/100)

Z = (0.74 – 0.80) / 0.04

Z = -0.06/0.04

Z = -1.5

P-value = 0.133614

P-value > = 0.01

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that proportion is consistent with the manufacturer’s claim.

C- If you want to estimate the difference to within 1.0%, what sample size should you use (99% confidence level)?

Answer:

We are given

E = 1% = 0.01

Confidence level = 99%

Critical Z value = 2.5758

(By using Z-table)

Sample size formula is given as below:

Sample size = n = (Z/E)^2*p*q

n = (2.5758/0.01)^2*0.80*0.20 = 10615.59

Required sample size = 10616

D- What is the observed significance level (p-value)?

Answer:

Observed significance level or p-value is given as below:

P-value = 0.133614

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