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
Answer:
(a) The probability that the store’s revenues were at least $9,000 is 0.0233.
(b) The revenue of the store on the worst 1% of such days is $7,631.57.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.
Then, the mean of the distribution of the sum of values of X is given by,
And the standard deviation of the distribution of the sum of values of X is given by,
It is provided that:
As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.
(a)
Compute the probability that the store’s revenues were at least $9,000 as follows:
Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.
(b)
Let <em>s</em> denote the revenue of the store on the worst 1% of such days.
Then, P (S < s) = 0.01.
The corresponding <em>z-</em>value is, -2.33.
Compute the value of <em>s</em> as follows:
Thus, the revenue of the store on the worst 1% of such days is $7,631.57.