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3 years ago

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Suppose we want a 90% confidence interval for the average amount spent on entertainment (movies, concerts, dates, etc.) by fresh

man in their first semester at a large university. The interval is to have a maximum bound on the error (or simply margin of error) of $2, and the amount spent has a normal distribution with a known standard deviation $30. The number of observations required is at least

1 answer:

katovenus [111]3 years ago

6 0

Answer:

609

Step-by-step explanation:

Standard deviation = $\sigma$ = $30

Margin of error = E = $2

Confidence level = 90%

Since the distribution is said to be normal, we will use z scores to solve this problem.

The z score for 90% confidence level = z = 1.645

Sample size= n = ?

The formula to calculate the margin of error is:

$E=z\frac{\sigma}{\sqrt{n}}\\\\\sqrt{n}=z\frac{\sigma}{E}\\\\n=(\frac{z\sigma}{E} )^{2}$

Using the values in above equation, we get:

$n=(\frac{1.645 \times 30}{2} )^{2}\\\\ n = 608.9$

This means, the minimum number of observations required is 609

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None of the expression are equivalent to $49^{(2t - 0.5)}$

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Given

$49^{(2t - 0.5)}$

Required

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$49^{(2t - 0.5)}$

Expand 49

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Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

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$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

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Expand numerator

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