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Answer:
A sample of 997 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
The margin of error is of:
A previous study indicates that the proportion of left-handed golfers is 8%.
This means that
98% confidence level
So , z is the value of Z that has a p-value of
, so
.
How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?
This is n for which M = 0.02. So
Rounding up:
A sample of 997 is needed.
Using the <u>normal distribution and the central limit theorem</u>, it is found that there is a 0.0166 = 1.66% probability of a sample proportion of 0.59 or less.
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sampling proportions of a proportion p in a sample of size n has mean
and standard error
In this problem:
- 1,190 adults were asked, hence
- In fact 62% of all adults favor balancing the budget over cutting taxes, hence
.
The mean and the standard error are given by:
The probability of a sample proportion of 0.59 or less is the <u>p-value of Z when X = 0.59</u>, hence:
By the Central Limit Theorem
has a p-value of 0.0166.
0.0166 = 1.66% probability of a sample proportion of 0.59 or less.
You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213