For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
1 answer:
You might be interested in
Answer:
We conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.
Step-by-step explanation:
We are given that the company claims that more than 90% of all orders are mailed within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid.
A recently taken sample of 175 orders showed that 161 of them were mailed within 72 hours.
<u><em>Let p = percentage of all orders that are mailed within 72 hours.</em></u>
SO, Null Hypothesis, : p
90% {means that less than or equal to 90% of all orders are mailed within 72 hours after they are received}
Alternate Hypothesis, : p > 90% {means that more than 90% of all orders are mailed within 72 hours after they are received}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = proportion of orders that were mailed within 72 hours in a sample of 175 =
= 92%
n = sample of orders = 175
So, <u><em>test statistics</em></u> =
= 0.975
The value of the test statistics is 0.975.
<em>Now at 1% significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is less than the critical value of z as 0.975 < 2.3263, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>
Therefore, we conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.
Answer is below........
Answer:
a) The 95% confidence interval for the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.
b) The confidence interval is built from a sample of 48 days. However, we can be 95% sure that for all days, the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution 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 = 48 - 1 = 47
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 47 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0117
The margin of error is:
M = T*s = 2.0117*2.3 = 4.6
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 7.1 - 4.6 = 2.5 hours of sunlight per day
The upper end of the interval is the sample mean added to M. So it is 7.1 + 4.6 = 11.7 hours of sunlight per day
a) Compute the 95% confidence interval for the mean.
The 95% confidence interval for the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.
b) Explain your confidence interval from part (a) in context.
The confidence interval is built from a sample of 48 days. However, we can be 95% sure that for all days, the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.