What is 3425030 in scientific notation
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Answer:
The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution should be used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 28 - 1 = 27
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.314
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.6 - 0.646 = $33.954
The upper end of the interval is the sample mean added to M. So it is 34.6 + 0.646 = $35.246
The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.