Easy question: y = -8x What is x if you substitute 3 for y?
1 answer:
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Answer: No, New York students are not truly more nerdy than the California students.
Step-by-step explanation:
Since we have given that
Number of students in a group = 50
Mean score of New York psychology = 530
Mean score of California psychology = 515
Standard deviation = 80
So, we will find t-distribution first.
Let α = 5% = 0.05
So,
Since t < t(critical value)
No, New York students are not truly more nerdy than the California students.
Answer:
A) See the picture
B) 14
C) 45%
Step-by-step explanation:
A) To create a histogram like the one on the picture you can use an online tool like socscistatistics where the number of classes is customizable
B) Because the question B and C have to be responded using a frequency table with 8 classes the answer is 14; the method of using cumulative frequency tables should only be considered as a way of estimation, that is because you obtain values that depend on your choice of class intervals. The way to get a better answer would be to use all the scores in the distribution
Pc1 = 100*(4/40) = 10
Pc2 = 100*(4/40) = 10
Pc3 = 100*(3/40) = 7.5
Pc4 = 100*(11/40) = 27.5
Pc5 = 100*(5/40) = 12.5
Pc6 = 100*(4/40) = 10
Pc7 = 100*(7/40) = 17.5
Pc8 = 100*(2/40) = 5
Pc8 + Pc7 + Pc6 + Pc5 + Pc4 + Pc3 + Pc2 = 90%
Therefore, From class 8 to class 2 is the top 90% of the applicants and the minimum score is 14.
C) Scores equal to or greater than 20 are from class 8 to class 5
Pc8 + Pc7 + Pc6 + Pc5 = 45%
Answer:
(a) Null Hypothesis, : p = 0.50
Alternate Hypothesis, : p > 0.50
(b) The value of level of significance (α) given in the question is 0.10.
(c) The sampling distribution of the sample statistic is Normal distribution.
(d) This test is right-tailed.
(e) The value of z test statistics is 0.96.
(f) The P-value is 0.1685.
(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.
(h) We conclude that the proportion of baby girls is equal to 0.50.
Step-by-step explanation:
We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5.
Assume that sample data consists of 78 girls in 144 births.
Let p = <u><em>population proportion of baby girls</em></u>
(a) So, Null Hypothesis, : p = 0.50 {means that the proportion of baby girls is equal to 0.50}
Alternate Hypothesis, : p > 0.50 {means that the proportion of baby girls is greater than 0.50}
(b) The value of level of significance (α) given in the question is 0.10.
(c) The sampling distribution of the sample statistic is Normal distribution.
(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.
(e) The test statistics that would be used here <u>One-sample z test for proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of baby girls =
= 0.54
n = sample of births = 144
So, <u><em>the test statistics</em></u> =
= 0.96
The value of z test statistics is 0.96.
(f) <u>The P-value of the test statistics is given by;</u>
P-value = P(Z > 0.96) = 1 - P(Z < 0.96)
= 1 - 0.8315 = 0.1685
<u></u>
(g) <u>Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.</u>
(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that the proportion of baby girls is equal to 0.50.