Question

Universidad Ana G Méndez

Recinto Gurabo

Asignación #4

Cyd Mary Alicea Cora

#S01011727

Prof: Juan Valera

Stat-555

Solve each problem.

1. Boys of a certain age are known to have a mean weight of μ = 88 pounds. A complaint is

made that the boys living in a municipal children's home are underfed. As one bit of

evidence, n = 36 boys (of the same age) are weighed and found to have a mean weight

of x¯ = 81.94 pounds. It is known that the population standard deviation σ is 12 pounds.

Based on the available data, what should be concluded concerning the complaint?

Ho: μ = 88

Ha: μ < 88

Z= (x- μ) / (σ/sqrt (n))

= (81.94-88) / (12/sqrt(36))

= -3.03

Critical value at 5% is Z0.05 =1.645

-3.03 < -1.645, podemos rechazar a Ho y así concluir que tenemos pruebas

suficientes para respaldar la queja formulada. De que los niños que viven en un

hogar municipal de niños están infestados.

2. It is assumed that the mean systolic blood pressure is μ = 118 mm Hg. In the Caguas

Heart Study, a sample of n = 81 people had an average systolic blood pressure of 127

mm Hg with a standard deviation of 20 mm Hg. Is the group significantly different (with

respect to systolic blood pressure!) from the regular population? (Remember that σ is

unknown).

3. The president of a large electric utility claims that 85 percent of his 2,000,000 customers

are very satisfied with the service they receive. To test this claim, the local newspaper

surveyed 200 customers, using simple random sampling. Among the sampled customers,

77 percent say they are very satisfied. Based on these findings, can we reject the

president's hypothesis that 85% of the customers are very satisfied? Use a 0.05 level of

significance.

4. A rental car company claims the mean time to rent a car on their website is 65 seconds

with a standard deviation (σ) of 30 seconds. A random sample of 49 customers

attempted to rent a car on the website. The mean time to rent was 77 seconds. Is this

enough evidence to contradict the company's claim?

5. According to the ACME, the mean height of adults ages 21 and older is about 67 inches.

Let's test if the mean height of our sample data is significantly different than 66.5 inches

using a one-sample t test.

Our sample is:

x=c(61,72,75,70,63,69,73,60,59,69,67,64,65,71,70)

Answer 1

Solution:-

5)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 66.50
Alternative hypothesis: u 66.50

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.280625

DF = n - 1

D.F = 14
t = (x - u) / SE

t = 0.547

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 14 degrees of freedom is less than -0.547 or greater than 0.547.

Thus, the P-value = 0.593

Interpret results. Since the P-value (0.593) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the mean height of our sample data is significantly different than 66.5 inches.