86 divide by 3 using area model,
1 answer:
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Answer:
We conclude that the daily average revenue was actually $675.
Step-by-step explanation:
We are given that the current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75.
A sample of 30 days reveals a daily average revenue of $625.
<u><em>Let </em></u><u><em> = daily average revenue.</em></u>
So, Null Hypothesis, :
= $675 {means that the daily average revenue was $675}
Alternate Hypothesis, :
$675 {means that the daily average revenue was different from $675}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample daily average revenue = $625
= population standard deviation = $75
n = sample of days = 30
Since, we are given that we have decided not to reject the null hypothesis which leads us to the conclusion that the daily average revenue was actually $675.
Answer:
It is concluded that no difference exists in the comprehension of the lab based on the test scores.
Step-by-step explanation:
<em><u>Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.</u></em>
The above statement tells us that the independent t or two sample t test will be performed as both have equal variances and are normally distributed.
This can be easily done through excel.
The following table is obtained
t-Test: Two-Sample Assuming Equal Variances
<u> Text Visual Illustrations</u>
Mean 70.28 75.08
Variance 304.48 228.58
Observations 15 15
Pooled Variance Sp²= 266.53
Pooled Standard Deviation = Sp = 16.33
Hypothesized Mean Difference = x1`-x2`= 0
df = n1+n2-2= 15+15-2= 30-2= 28
t Stat -0.805188239
<u>P(T<=t) two-tail 0.427495979 </u>
<u>t Critical two-tail 1.701130908 </u>
Let the null and alternate hypotheses be
H0 : u1-u2= 0 against the claim Ha: u1-u2≠0
There is no difference between the means
against the claim
that there is a difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.
The significance level is ∝= 0.1
The d.f is n1+n2-2= 15+15-2= 30-2=28
This is a two tailed test and the critical region is t (0.025) (28) ≥ 1.7011 and t (0.025) (28) ≤ - 1.7011.
The test statistic is
t= x1-x2/ Sp √1/n1+ 1/n2
t= 70.28 -75.08/ 16.33√1/15 +1/15
t= -4.8/5.963
t= -0.8049811 ( minute difference from excel result due to rounding)
Since the calculated value of t= -0.8049 does not fall in the critical region we conclude that there is no difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.
We accept the null hypothesis.
The p- value is 0.427495979.
