Question 1193764: A data set about speed dating includes like" ratings of male dates made by the female dates. The summary statistics are n = 191 , x = 5.83 , s = 1.98 . Use a 0.05 significance level to test the claim that the population mean of such ratings is less than 6.00. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
mu = population mean of the rating
Claim: "The population mean of such ratings is less than 6.00"
Translation into symbols: mu < 6.00
The goal of this entire solution is to test if this claim is true or not.
Hypotheses:
H0: mu = 6.00
H1: mu < 6.00
Claim is in the alternative hypothesis H1
We have a left-tailed test because of the inequality sign here.
Some stats textbooks would state the null as mu ≥ 6.00, but I'll stick with mu = 6.00 so that we focus on exactly one mu value only. The null must be centered around one single value.
Given info:
n = 191 = sample size
xbar = 5.83 = sample mean
s = 1.98 = sample standard deviation
Test Statistic:
z = (xbar - mu)/( s/sqrt(n) )
z = (5.83 - 6.00)/( 1.98/sqrt(1.91) )
z = -0.11865892643357
z = -0.12
Despite sigma being unknown, we can use z here because n > 30 is true. The student T distribution is very similar to the standard normal Z distribution for n > 30.
Use a calculator like this
https://onlinestatbook.com/2/calculators/normal_dist.html
to find that P(Z < -0.12) = 0.4522 approximately
If you have your TI calculator with you and wish to use it, then hit the key labeled "2ND" and hit the "VARS" button to bring up the stats menu. The function you want is called normalcdf.
You can type in something like normalcdf(-99, -0.12) and you should get some value close to 0.4522
Alternatively, you can use a Z table like this
https://www.ztable.net/
to find that P(Z < -0.12) = 0.45224 approximately.
Look at the row that starts with -0.1 and the column that has 0.02 up top. The two combine to have 0.45224
Note: We shade to the left of z = -0.12 directly because we're doing a left-tailed test.
The p-value is roughly 0.4522
It is the probability of selecting a z score of z = -0.12 or smaller. It is also the probability of getting xbar = 5.83 or smaller.
Comparing the p-value (0.4522) and the alpha value (0.05), we see the p-value is clearly much larger.
Therefore, we fail to reject the null. We only reject the null when the p-value is smaller than alpha.
It is said we "accept" the null. I put that in quotes because we don't really accept it, rather we just don't have enough data to overturn it.
Side note: The alpha level is the probability of a Type I error.
Since we failed to reject the null, we have no choice but to "accept" that mu = 6.00 is the case.
To address the original claim, we have shown it to be false because mu is not less than 6.00
Recall the claim was made in the alternative hypothesis.
We don't have enough statistically significant evidence to overturn the null and side with the alternative.
Yes the xbar = 5.83 is smaller than 6.00, but this appears to be due to random chance.
If you were to do another random sample of 191, then you might get some xbar equal to 6.00 or larger.
----------------------------------------------------
Summary:
Null hypothesis: H0: mu = 6.00
Alternative hypothesis: H1: mu < 6.00
Test Statistic: z = -0.12
P-value = 0.4522
Decision: Fail to reject the null (i.e. "accept" the null)
Conclusion: The population mean of the ratings is not less than 6.00 (thereby the claim made at the top is false)
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