What is the total price of a $45.79 item when 7% sales tax is added?
2 answers:
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Answer:
we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years
Step-by-step explanation:
Given that in a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years.
Let X be the group married less than 2 years and Y less than 5 years
X Y Total
Sample size 300 300 600
Favouring 240 288 528
p 0.8 0.96 0.88
p difference = -0.16
Std error for difference =
Test statistic = p difference/std error=-6.03
p value <0.000001
Since p is less than alpha 0.05 we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years
Answer:
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears
Step-by-step explanation:
For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.
However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears
This probability is the pvalue of Z when X = 82. So
has a pvalue of 0.0885.
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears