Arrange the data in equal classes of size 10 starting from 40 percent.
Frequency table
From the table, we compute the sample mean "\\bar{X}" as follows
Also, the sample variable is calculated as follows:
"Var(X)={ 1\\over 100-1}\\bigg(422100-{6410^2 \\over 100}\\bigg)\\approx113.3232"
Therefore, the sample sandartd deviation "s" is directly computed by taking the square root from the sample variance.
a) The mean class is "60-70."
b) Find 95 % confidence interval
"CI=\\bar{X}\\pm z_c\\cdot{s \\over \\sqrt{n}}""CI=64.1\\pm 1.96\\cdot{10.6453 \\over \\sqrt{100}}""CI=64.1\\pm 2.086"
Find 99 % confidence interval
"CI=\\bar{X}\\pm z_c\\cdot{s \\over \\sqrt{n}}"
"CI=64.1\\pm 2.5758\\cdot{10.6453 \\over \\sqrt{100}}"
"CI=64.1\\pm 2.742"
"(61.358, 66.842)"
We can be 99% confident that the population mean (μ) falls between 61.358 and 66.842.
Since "66.842<72", then we would we be right to say that this group of candidates performed below the average.
c) A sample of 50 scores were obtained from 2011 series of a similar examination. The sample gave a mean score of 66.7 percent with a standard deviation of 3 percent.
Would we be right to conclude that there was no difference in performance between 2011 and 2012 series of the examination?
The provided sample means are shown below:
"\\bar{X_1}=64.1, \\bar{X_2}=66.7"
Also, the provided population standard deviations are:
"\\sigma_1=10.6453, \\sigma_2=3"
and the sample sizes are "n_1=100, n_2=50,"
The following null and alternative hypotheses need to be tested:
"H_0:\\mu_1=\\mu_2"
"H_1: \\mu_1\\not=\\mu_2"
This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.
Rejection Region
The significance level is "\\alpha=0.05," and the critical value for a two-tailed test is "z_c=1.96"
The rejection region for this two-tailed test is "R=\\{z:|z|>1.96\\}"
Test Statistics
The z-statistic is computed as follows:
"z={64.1-66.7 \\over \\sqrt{10.6453^2\/100+3^2\/50}}=-2.269"
Since it is observed that "|z|=2.269>z_c=1.96," it is then concluded that the null hypothesis is rejected.
Using the P-value approach:
"p-value=2\\cdot0.011634=0.023268"
The p-value is "p=0.0233," and since "0.0233<0.05," it is concluded that the null hypothesis is rejected.
We would not be right to conclude that there was no difference in performance between 2011 and 2012 series of the examination.
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