a )
H0: p = 0.08
Ha: p
0.08
b)
sample proportion
= 21 / 194 = 0.1082
c)
Test statistics
z = (
- p) / SE
= (0.1082 - 0.08) / 0.0195
= 1.45
d)
p-value = 2 * P(Z > z)
= 2 * P(Z > 1.45)
= 2 * 0.0735
= 0.1470
e)
Since p >=
, we do not have enough evidence to reject the null
hypothesis.
It is believed that nearsightedness affects about 8% of all children. In a random sample of...
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It is believed that nearsightedness affects about 8% of all children. In a random sample of 200 children, 25 are nearsighted. Do these data provide evidence that the 8% value is inaccurate? At α = 0.05, test the claim. Find the p-value. P-value = 0.0190 P-value = 2.35 P-value = 0.05 P-value = 0.125
It is generally believed that nearsightedness affects about 15% of children. A school district gives vision tests to 111 incoming kindergarten children. In our sample of 111 students, we find 13% of the students were nearsighted. Construct a 90% confidence interval for the number of nearsighted kindergarteners we would expect to see based on our sample. Does this support or refute the estimate of 15%? Assume conditions are met (so don't check them)! Now that you have your interval, we...
6.16 Is college worth it? Part I: Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school. (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey...
Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school. (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to...
It is generally believed that nearsightedness affects about 15% of children. A school district gives vision tests to 111 incoming kindergarten children. Use the empirical rule (68%-95%-99.7% Rule) to determine what proportion of nearsighted children we might expect to see in samples of 111 children (I'm not looking for the number of children). Assume conditions are met! Based on your results, would you be surprised to find a sample where 20% of children were nearsighted? Find the z-score and resulting probability to...
It is generally believed that nearsightedness affects about 15% of children. A school district gives vision tests to 111 incoming kindergarten children. In our sample of 111 students, we find 13% of the students were nearsighted. Construct a 90% confidence interval for the number of nearsighted kindergarteners we would expect to see based on our sample. Does this support or refute the estimate of 15%? Assume conditions are met (so don't check them)! Explain in details
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