# 8.1 A single population mean using the normal distribution  (Page 5/20)

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## Try it

Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.

(33.37, 38.63)

Suppose we change the original problem in [link] to see what happens to the error bound if the sample size is changed.

Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36?

• $\overline{x}$ = 68
• EBM = $\left({z}_{\frac{\alpha }{2}}\right)\left(\frac{\sigma }{\sqrt{n}}\right)$
• σ = 3; The confidence level is 90% ( CL =0.90); ${z}_{\frac{\alpha }{2}}$ = z 0.05 = 1.645.

## Solution b

If we decrease the sample size n to 25, we increase the error bound.

When n = 25: EBM = $\left({z}_{\frac{\alpha }{2}}\right)\left(\frac{\sigma }{\sqrt{n}}\right)$ = (1.645) $\left(\frac{3}{\sqrt{25}}\right)$ = 0.987.

## Summary: effect of changing the sample size

• Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
• Decreasing the sample size causes the error bound to increase, making the confidence interval wider.

## Try it

Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.

(34.6041, 37.3958)

## Working backwards to find the error bound or sample mean

When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.

## Finding the Error Bound

• From the upper value for the interval, subtract the sample mean,
• OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two.

## Finding the Sample Mean

• Subtract the error bound from the upper value of the confidence interval,
• OR, average the upper and lower endpoints of the confidence interval.

Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.

Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean.

## Calculate the error bound:

• If we know that the sample mean is 68: EBM = 68.82 – 68 = 0.82.
• If we don't know the sample mean: EBM = $\frac{\left(68.82-67.18\right)}{2}$ = 0.82.

## Calculate the sample mean:

• If we know the error bound: $\overline{x}$ = 68.82 – 0.82 = 68
• If we don't know the error bound: $\overline{x}$ = $\frac{\left(67.18+68.82\right)}{2}$ = 68.

## Try it

Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.

Sample mean is 45, error bound is 2.88

## Calculating the sample size n

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.

binomial distribution tends to normal distribution
if sample size n very large and probability tends to 0.5 if these both conditions are satisfied by binomial distribution it would tends to normal distribution
sravani
n tends to infinite i.e large Probability tends to 0 i.e indefinitely small. np = lamda
Anji
the above is poison to Bin
Anji
no of trails n tends to indefinitely large..i.e infine neither p nor q is very small Then bin tends to normal
Anji
if the death of of the snow is my yard is normally distributed with the m is equals to 2.5 and what is the probability that a randomly chosen location with have a no that between 2.25 and 2.76
hey
Shubham
🤔
Iqra
hello
Sakshi
hii
Rushikesh
helow
why Statistics so hard
Mohd
ho geya solve
Sakshi
it's not hard
Sakshi
it is hard 😭
Mohd
solution?
Abdul
hii
it's just need to be concentrate
Akinyemi
exactly..... concentration is very important
Iqra
rewrite the question
what is the true statement about random variable?
A consumer advocate agency wants to estimate the mean repair cost of a washing machine. the agency randomly selects 40 repair cost and find the mean to be $100.00.The standards deviation is$17.50. Construct a 90% confidence interval for the mean.
pls I need understand this statistics very will is giving me problem
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
No
Ariel
how do I find the modal class
look for the highest occuring number in the class
Kusi
the probability of an event occuring is defined as?
The probability of an even occurring is expected event÷ event being cancelled or event occurring / event not occurring
Gokuna
what is simple bar chat
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
statistics is percentage only
the first word is chance for that we use percentages
it is not at all that statistics is a percentage only
Shambhavi
I need more examples
how to calculate sample needed
mole of sample/mole ratio or Va Vb
Gokuna
how to I solve for arithmetic mean
Yeah. for you to say.
James
yes
niharu
how do I solve for arithmetic mean
niharu
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ... *2sinQ/Cos 3Q = tan 3Q - tan Q