a journal reports that 34 percent of for child An article in SELECT ALL APPLICAB

Question

a journal reports that 34 percent of for child An article in SELECT ALL APPLICABLE CHOICES fathers take no responsibility A) There is enough evidence to support the Claim B) Not enough evidence to reject the Claim care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 234 fathers from Littleton yielded 96 who did not help with child carc. Test the rescarcher's claim at the 0.05 significance level. (there are three correct choices in this question) D) Ho: p2 34 ps 34 H1: p .34(claim) E) There is enough evidence to reject the Ho F There is enough evidence to reject the Claim G) Not enough evidence to reject the Ho H) There is enough evidence to support the Ho I) None of thesc The claim is that the proportion of drowning dcaths of childrcn attributablc to bcaches is more than 0.25, and the sample statistics include n = 575 drowning deaths of children with 30 percent of them attributable to beaches.. Find the value of the test statistic z using -an SELECT ALL APPLICABLE CHOICES A) -2.77 B) 2.62 C) 2.77 D) -2.62 Pp E) None of thesc that a simple random sample has been SELECT ALL APPLICABLE CHOICES selected from a normally distributed population A) There is enough evidence to reject the Ho and test the given claim. State the final conclusion that addresses the original claim andC select three correct choices. Usc a significance level of = 0.05to test the claim that = 32.6 The sample data consist of 15 scorcs for which B) There is enough evidence to support the Claim D) Not enough evidence to reject the E)Not enough evidence to reject the Ho Claim H0 : =32.6 (claim) F) There is enough evidence to reject the Claim 1: 32.6 41.6and 8 8 Use the traditional method of testing hypotheses. H0 : (claim) >32.6

Explanation / Answer

11)

We have to test the hypothesis that,

H0 : p < 0.34 Vs H1 : p > 0.34

where p is the population proportion.

Assume alpha = level of significance = 0.05

Given that,

x = 96

n = 234

Here test statistic follows Z-distribution.

The test statistic is,

Z = (p^ - p) / sqrt[(p*q) / n]

where p^ is sample proportion.

p^ = x/n = 96/234 = 0.41

q = 1 - p = 1 - 0.34 = 0.66

Z = (0.41 - 0.34) / sqrt [(0.34*0.66) / 234 ]

Z = 2.27

Now we have to find P-value for taking decision.

P-value we can find in EXCEL.

syntax :

=1 - NORMSDIST(z)

where z is test statistics

P-value = 0.012

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that the population proportion is greator than 0.34

12) The hypothesis for the test is,

H0 : p < 0.25 Vs H1 : p > 0.25

x = 30

n = 575

p^ = 30/575 = 0.05

q = 1 - 0.25 = 0.75

z = (0.05 - 0.25) / sqrt [(0.25*0.75) / 575 ] = -10.96

Therefore answer is none of these.

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