Question

6.3.3 Marks on an exam in a statistics course are assumed to be normally distributed

Answer 1

Answer:

- The calculated p-value (0.392452) is higher than the significance level at which the test was performed, hence, the null hypothesis is true and μ = 60

- 95% Confidence interval for the population mean score = (47.4, 84.1)

Explanation:

The sample of 4 students had scores of 52, 63, 64, 84.

First of, we need to compute the sample mean, we do not need the sample standard deviation as the population variance is given as 5

Mean = (Σx)/N

x = each variable

N = number of variables = 4

Mean = (52 + 63 + 64 + 84)/4

Mean = 65.75

Sample Standard deviation = σ

= √[Σ(x - xbar)²/N]

xbar = mean = 65.75

Σ(x - xbar)² = 532.75

σ = √[532.75/4] = 11.54

in hypothesis testing, the first thing is usually to state the null and alternative hypothesis.

From the question, the null hypothesis has already been stated as

H₀: μ = 60

The alternative hypothesis would then be that the population mean score isn't equal to 60

Hₐ: μ ≠ 60

Since the population distribution is normal and the sample standard deviation is to be used, we use the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 65.75

μ₀ = Standard to be compared against = 60

σₓ = standard error = (σ/√n) = (11.54/√4) = 5.77

t = (65.75 - 60)/5.77 = 0.9965 = 1.00

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level = 0.05 (95% confidence level)

The hypothesis test uses a two-tailed condition because we're testing in two directions.

p-value (for t = 1.00, at 0.05 significance level, df = 3, with a two tailed condition) = 0.392452

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.392452

0.392452 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there is enough evidence to conclude that the populatiom mean score is equal to 60.

b) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 65.75

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 3) = 3.18 (from the t-tables)

Standard error of the mean = 5.77

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 65.75 ± (3.18 × 5.77)

CI = 65.75 ± 18.3486

95% CI = (47.4014, 84.0986)

95% Confidence interval = (47.4, 84.1)

Hope this Helps!!!