A rectangle has width of 2x2

According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

Let $\hat p$ = sample proportion of Americans who can order a meal in a foreign language

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = 0.19766

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0

3 years ago

The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they

sukhopar [10]

Answer:

From the result of the one-tailed z-test, we can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Step-by-step explanation:

Sample size = 50

Mean waiting time = 13 minutes

Standard deviation = 11 minutes

We need to perform one hypothesis test that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

The null hypothesis would be that there is no significant difference.

H₀: μ₀ = 15 minutes

The alternative hypothesis would be that the mean waiting time is indeed significantly less than the 15-minute claim by the taxpayer advocate.

Hₐ: μ₀ < 15 minutes.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim). Hence, we can use the z-test

z = (x - μ)/σₓ

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15)/1.556 = - 1.29

Using the z-table at significance level of 0.05.

p-value = 1 - 0.9115 (from the z-tables)

p-value = 0.0885

Since the p-value is more than the significance level (0.0885 > 0.05), we do not reject the null hypothesis.

Hence, we accept the null hypothesis and can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Hope this Helps!!!

4 0

3 years ago

Help me please........

Kobotan [32]

False should be the answer

8 0

3 years ago

Three friends went to a fundraiser at a local restaurant. Each plate cost $7.95, which included a drink. A tip of 20% was left f

miskamm [114]

Answer:

159 try it