The positive value of Z is 2.17.
What do you mean by normal distribution?
A data collection with a normal distribution is put up so that the majority of the values cluster in the middle of the range and the remaining values taper off symmetrically in either direction. Because of its flared appearance, a normal distribution's graphic representation is occasionally referred to as a bell curve. The specific shape may change depending on how the population is distributed, but the peak and curve are always symmetrical and in the centre.
The information can be represented in the standard normal curve below:
It is given that 97% of the standard normal curve lies between −z and z.
The area in between is 0.97 meaning that one side is (0.97)/2 = 0.485
In addition, the region not shaded is the level of significance represented by (1-0.97)/2 = 0.03/2 = 0.015
From the Z-Score table (0 to Z table), this probability corresponds to the Z value of Z= 2.17 and -2.17
Thus the positive value of Z is 2.17.
To learn more about the normal distribution from the given link.
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Answer: c. -2.15
Step-by-step explanation:
As per given we have:
Null hypothesis :
Alternative hypothesis :
let p be the proportion of firms in the manufacturing sector offer child-care benefits
A random sample of 390 manufacturing firms is selected and 295 of them offered child-care benefits.
sample size : n= 390
sample proportion :
Test statistic for proportion :
Substitute the values , we get
Hence, the value of test statistic = -2.15
Correct option is c. -2.15 .
Let's call the height of the tower and the distance from the 65 degree angle to the base of the tower
.
We get two equations:
Answer: 48
Answer:
Step-by-step explanation:
Hello!
A study was conducted to test the effect of classical music on the brain. For this 20 6th grade students were randomly divided into two independent groups of 10, the same math test was given to these students. The first group listened to classical music for 20 min before taking the test. The second group took the test without listening to music.
Sample 1 (With music)
X₁: Score of a 6th grade student that heard 20 min classical music before taking the math test.
91 77 58 89 83 78 74 81 91 88
n₁= 10
X[bar]₁= 81
S₁= 10.11
Sample 2 (Without music)
X₂: Score of a 6th grade student that didn't hear classical music before taking the math test.
81 65 69 69 67 61 67 87 64 81
n₂= 10
X[bar]₂= 71.10
S₂= 8.70
Asuming both variables have a normal distribution and the population variances are unknown but equal, the statistic to use for both the CI and hypothesis tests is:
t= (X[bar]₁-X[bar]₂) - (μ₁ - μ₂) ~t
Sa
a) 95% for (μ₁ - μ₂)
(X[bar]₁-X[bar]₂) ± *Sa
Sa²= 88.94
Sa= 9.4308 ≅ 9.43
(81-71.10) ± 2.101*(9.43*)
[1.04;18.76]
With a confidence level of 95% youd expect that the interval [1.04;18.76] will contain the difference between the population means of the test scores of the kids that listened to classic music and the kids that didn't listen to music before taking the math test.
b)
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
α: 0.05
One-tailed test (right tail)
Critical value
Rejection region t ≥ 1.734
t= (81-71.10) - 0 = 2.34
9.43*
The decision is to reject the null hypothesis.
c) You are asked to conduct the same test at a different levelm this means that only the significance level changes:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
α: 0.01
One-tailed test (right tail)
Critical value
Rejection region t ≥ 2.552
t= (81-71.10) - 0 = 2.34
9.43*
The decision is to not reject the null hypothesis.
At a significance level of 5%, the decision is to reject the null hypothesis, which means that the population average of the test scores of 6th-grade children that listened to classical music before taking a math test is greater than the population average of the test scores of 6th graders that took the math test without listening to classical music.
But at 1% significance level, there is not enough evidence to reject the null hypothesis. At this level, the conclusion is that the average test score of 6th graders that listened to classical music before taking the math test is at most equal to the average test score of 6th graders that didn't listen to music before the test.
I hope it helps!