Question

Write out the null and alternative hypotheses for the following hypothetical proposal. Carry out a one-sample Z test to determine significance at the α=0.05 level.

PROPOSAL

Recent scientific literature has suggested a link between exposure to the endocrine disruptor Bisphenol A (BPA) found ubiquitously in many plastic materials, and precocious puberty in adolescents. The transmission is maternal to child during the last weeks of gestation. Scientists can trace the presence of BPA in children for years after birth, and believe that it may be the cause for early onset of puberty in young girls. Suppose a researcher collects a SRS of 92 girls with high exposure rates of BPA and follows them through puberty. The sample average age of puberty was 9.7 years. Assume the national average age for onset of puberty is 11.2 years with a population standard deviation of 0.6 years. Does the researcher have sufficient evidence at the alpha level of 0.05 to determine if the exposure to BPA has led to a decreased age of puberty in girls?

option 1 Ho: µ=11.2, Ha: µ>11.2 Z stat=44.81 pvalue of <0.001 Fail to reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have not reached puberty at a significantly lower age. |
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option 2 Ho: µ=11.2, Ha: µ<11.2 Z stat=23.98 pvalue of <0.0001 Reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have reached puberty at a significantly lower age. |
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option 3 Ho: µ=11.2, Ha: µ<11.2 Z stat=30.21 pvalue of <0.005 Reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have reached puberty at a significantly lower age. |

Answer #1

This is the left tailed test .

The null and alternative hypothesis is

H_{0} :
= 11.2

H_{a} :
< 11.2

Test statistic = z

= ( - ) / / n

= (9.7 - 11.2) / 0.6 / 92

Test statistic = 23.98

P-value = 0.0000

P-value <

Reject the null hypothesis .

The correct option is :

**option 2**

Ho: µ=11.2, Ha: µ<11.2

Z stat=23.98 pvalue of <0.0001 Reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have reached puberty at a significantly lower age.

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