A random sample of 850 births included 434 boys. Use a 0.10 significance level to test the claim that 51.5% of newborn babies are boys. Do the results support the belief that 51.5% of newborn babies are boys? Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0: pequals0.515 Upper H 1: pnot equals0.515 Your answer is correct. B. Upper H 0: pequals0.515 Upper H 1: pgreater than0.515 C. Upper H 0: pnot equals0.515 Upper H 1: pequals0.515 D. Upper H 0: pequals0.515 Upper H 1: pless than0.515 Identify the test statistic for this hypothesis test.

A

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

z = -0.257

P value = P(Z<-0.257) = 0.797

Decision; we fail to reject the null hypothesis. That is, the results support the belief that 51.5% of newborn babies are boys

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) reject Null hypothesis

Z score < Z(at 90% confidence interval) accept Null hypothesis

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above;

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

Given;

n=850 represent the random sample taken

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 850

po = Null hypothesized value = 0.515

p^ = Observed proportion = 434/850 = 0.5106

Substituting the values we have

z = (0.5106-0.515)/√(0.515(1-0.515)/850)

z = −0.256677

z = −0.257

To determine the p value (test statistic) at 0.10 significance level, using a two tailed hypothesis.

P value = P(Z<-0.257) = 0.797

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = -0.257) which falls with the region bounded by Z at 0.10 significance level. And also the one-tailed hypothesis P-value is 0.797 which is greater than 0.10. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 10% significance level the null hypothesis is valid.

2y+y=5

add 2y and y