Identify the type I error and the type II error that correspond to the given hypothesis. The percentage of households with Inter
1 answer:
Answer:
Type I error : Option A
Type II error : Option C
Step-by-step explanation:
We are given that the percentage of households with Internet access is greater than 60%.
So, Null hypothesis, : p = 60%
Alternate Hypothesis, : p > 60%
Now, <u>Type I error states that</u> : Probability of rejecting null hypothesis given the fact that null hypothesis was true.
So, in our question Type I error will be Rejecting the null hypothesis that the percentage of households with Internet access is equal to 60% when that percentage is actually equal to 60%.
Now, <u>Type II error states that</u> : Probability of accepting null hypothesis given the fact that null hypothesis was false.
So, in our question Type Ii error will be Failing to reject the null hypothesis that the percentage of households with Internet access is equal to 60% when that percentage is actually greater than 60%.
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Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be .
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.
Amount of interest earned from lower-yielding account: .
Amount of interest earned from higher-yielding account: .
Let us solve for x.
Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be .
Therefore, the man invested $60,000 at 6%.