Previously 5% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of
1 answer:
Answer:
p value = 0.03514
Step-by-step explanation:
Hypotheses would be
(left tailed test at 10% level of significance)
Here p stands for the sample proportion of mothers smoked more than 21 cigarettes during their pregnancy.
Sample size =130
persons who smoked = 2
Sample proportion =
Assuming H0 to be true
Std error=
p difference = -0.0346
Test statistic z=-1.81
p value = 0.03514
Since p is less than 0.10, significance level, we reject H0
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Answer:
The null hypothesis is
The alternate hypothesis is
The test statistic is t = -1.95.
The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.
Step-by-step explanation:
Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.
At the null hypothesis, we test if the mean is 14 hours, that is:
At the alternate hypothesis, we test if the mean is less than 14 hours, that is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
14 is tested at the null hypothesis:
This means that
X = 13.6 hours, s = 1.3 hours. Sample of 40:
In addition to the values of X and s given, we have that
Test statistic:
The test statistic is t = -1.95.
P-value:
The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.
Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.