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Assignment - You just have to write the method how can calculate it.

# Interpret the result of the test of null hypothesis

# H0: "mean of population of x is less than 0.4"

# using these data and t.test():

set.seed (22)

x <-runif(15)

t.test(x,mu=0.4,altemative = 'greater')

# Output

# One Sample t-test

# data: x

# t = 3.9187, df = 14, p-value = 0.0007719

# alternative hypothesis: true mean is greater than 0.4

# 95 percent confidence interval:

# 0.5228421 Inf

# sample estimates:

# mean of x

# 0.6231297

Answer - Test Results

Here, we wish to determine whether the mean of X is greater than 0.4 and therefore, we perform one sample z test to test the claim and the direction of the test is one tailed test - right tailed test

The mean of x is 0.623. Here, we see that the mean of x for a sample of 15 is 0.623 with a standard deviation of 0.181. On comparing the mean values, we see that the mean of x for a sample of 15 is greater than the hypothesized value of 0.4.

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The sample information is given below

Sample Size = n = 15

Sample mean = x bar = 0.623

Sample standard deviation = s = 0.181

Population mean = µ = 0.4

Samples for this data can be selected using simple random sampling procedure. Simple random sampling is always high rated than the quota sampling technique due to its probability method used to collect the sample respondents. Samples selected using quota sampling will have high chance of biased results and hence simple random sampling is the best methods of selecting a representative sample from a population

Simple random sampling technique, is a probability sampling technique, in which, all samples have equal chance of being selected. For example, drawing a card numbered 10 from a pack of 52 cards, throwing a die and face turns up to 6, tossing a coin and getting head uses the simple random sampling techniques. This sampling technique is one of the best and easy sampling techniques and it is preferred by most researchers. This sampling technique is simple and do not involve any complex technique to select the samples from the desired population

Null Hypothesis: H₀: µ < = 0.4

That is, the mean of x is not greater than 0.4

Alternate Hypothesis: H_a: µ > 0.4 (Right tailed test)

That is, the mean of x is greater than 0.4

Level of Significance

Let the level of significance be α = 0.05

Rejection Region

If the p - value of t test statistic is greater than 0.05, then, we do not reject the null hypothesis at 5% level of significance.

If the p - value of t test statistic is less than 0.05, then, we reject the null hypothesis at 5% level of significance.

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Test Statistic

The t test statistic is calculated by using the formula given below

t=(x^- -μ)/(s⁄√n)=3.9187

Here, the value of t test statistic is 3.9187 and its corresponding p - value is 0.00077.

Since the p - value of t test statistic is less than 0.05, there is enough statistical evidence to reject the null hypothesis at 5% level of significance. Therefore, we can conclude that the mean of x is greater than 0.4 and statistically significant.

The 95% confidence interval for the mean of x is (0.5228, ∞). This indicates that when the samples are repeatedly taken from the same population, the, we can say that 95 out of 100 times the true mean population of x will fall within this interval.

The mean of x is 0.623. Here, we see that the mean of x for a sample of 15 is 0.623 with a standard deviation of 0.181. On comparing the mean values, we see that the mean of x for a sample of 15 is greater than the hypothesized value of 0.4.

Here, we see that the hypothesized value of 0.4 do not fall in this confidence interval and the lower limit of this confidence interval is greater than the hypothesized value of 0.4. Therefore, we can conclude that the mean of x is greater than 0.4 and statistically significant.

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