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Team TutorTeddy
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Question:
The Genetics and IVF Institute claim that the proportion of girls born to couples using its XSORT method is equal to 50%. In a recent study, among 920 babies born to couples using the XSORT method, 398 of the babies were girls. using these results and the 0.01 significance level, test the claim that the proportion of girls born to couples using it's XSORT method is equal to 50%.

1. State the claim in symbolic form:
P = 50

2. Identify the null and alternative hypotheses:
H0: p = .50
H1: p not equal .50

3. Calculate the value of the test statistic .... -4.43 (not 100% positive)

4. What is the p-value ...
.9999 (please confirm)

Answer: 3. p-hat = 398/920 = 0.4326

test statistics, z = (0.4326 - 0.50)/sqrt(0.50*(1 - 0.50)/920)
= -4.0886

4. P-value = P(z < -4.0886 ) + P(z > 4.0886) = 0.0

Question: Increasing numbers of businesses are offering childcare benefits for their workers. However, one union claims that more than 85% of firms in the manufacturing sector still do not offer any childcare benefits to their workers. A random sample of 250 manufacturing firms is selected, and only 30 of them offer childcare benefits. Specify the rejection region that the union will use when testing at a = 0.05.

Answer: Ho : p = 0.85
Ha : p > 0.85 (claim)

p-hat = (250 - 30)/250 = 0.88

z = (0.88 - 0.85)/sqrt(0.85*(1 - 0.85)/250) = 1.3284

critical z = 1.645 at alpha= 0.05 .

hence, computed z (1.3284) < critical z (1.645).
Fail to reject Ho.

conclusion : claim is not true.

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Question: In an old survey, 58% of adults say they have never heard of the internet. If 20 adults are selected at random, find the probability that exactly 12 will say they have never heard of the internet.

please explain how this is solved  

Answer: p = 0.58
n = 20
x = 12

P(x = 12) = 20c12*(0.58)^12*(1 - 0.58)^(20-12)
= 20!/(12!(20-12)!)(0.58)^12*(1 - 0.58)^8
= 0.1768 (Ans.)  

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Question: The time required for Dr. B's students to complete the stats exam is approx normally distributed with a MEAN of 40.4 minutes and standard deviation of 2.2 minutes. Let x be the random variable....

With the above settings find the probability that a student's exam completion time is at least 46 minutes.

a.0.0054
b.0.0183
c.0.9817
d.0.9946


Answer: P(x > = 46) = ?

z = (46 - 40.4)/2.2 =2.5454

P(x > = 46) = P(z > = 2.5454) = 0.0054 (Ans.)
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