Inferential Statistics help?

[匿名]

2017-10-20 4:38 pm

A manager wishes to determine whether the mean queuing time during peak hours for customers at a grocery store has changed from the target average queuing time of 4mins.
The manager tests this by sampling n = 40 customers at random and recording their queuing time. The average value of the sample is calculated to be 4.94 mins and queuing times in the sample vary by a standard deviation of s = 2.29 mins.
The hypotheses being tested are:
H0: μ = 4
Ha: μ ≠ 4.

Estimate the population mean,

The standard error,

the distribution,

The test statistic value TS,

Testing at a significance level a=0.05
the rejection region is;

less than

greater than
(2 decimal places)

is it/is not in the rejection region?
there is/is not evidence to reject the null hypothesis,

There is/is not sufficient evidence to suggest the average queuing time is different to 4 minutes..

Were any assumptions required in order for this inference to be valid?
a: No - the Central Limit Theorem applies, which states the sampling distribution is normal for any population distribution.
b: Yes - the population distribution must be normally distributed.
Insert your choice (a or b):

回答 (1)

cidyah

2017-10-20 7:59 pm

Estimate of the population mean = sample mean = 4.94

Standard error = σ/√n = 2.29/√40 = 0.36208

Sample mean = 4.94
Standard deviation = 2.29
Standard error of mean = s / √ n
Standard error of mean = 2.29 / √ 40
SE = 2.29/6.3246
Standard error of mean 0.3621
t = (xbar- μ ) / SE
t = (4.94-4) / 0.3621
t = 2.5961 (test statistic value TS)

P( |t| > 2.5961) = 0.0132
https://www.graphpad.com/quickcalcs/pvalue1.cfm

P-value < 0.05

It's in the rejection region
There is evidence to reject the null hypothesis

There is sufficient evidence to suggest the average queuing time is different to 4 minutes.

If the observations are normally distributed, we have a t-distribution

👍 0 👎 0