Statistical Analysis: Hypothesis Testing and One-Sample T-Test in SPSS

This assignment provides solutions to two statistical exercises involving hypothesis testing. The first exercise tests the hypothesis that the average number of gifts an NSW teacher receives per year is equal to 10, using a two-sided Z-test. The analysis includes outlier detection, normality tests, and the calculation of the Z-score, ultimately concluding that the null hypothesis cannot be rejected. The second exercise uses a one-sample t-test to determine if the average time a student takes to realize that the language being spoken is not English is greater than 60 seconds. The solution discusses how to handle outliers, performs normality tests, justifies the use of a one-sample t-test, and concludes that the average time is indeed greater than 60 seconds based on a significant p-value. The document also includes the assignment brief which provides context for the tutorials and exercises.

Running head: ASSIGNMENTS
Assignments
Name
Course Number
Date
Faculty Name

ASSIGNMENTS 2
Assignments
Assessment Exercise: Testing of Hypothesis
1. Hypothesis statements
Null hypothesis: The average number of gifts an NSW teacher receives per year = 10
Experimental hypothesis: The average number of gifts an NSW teacher receives per year ≠ 10
2. Decision criteria
This is a two-sided hypothesis test and we will reject the null hypothesis will be rejected if the
absolute calculated Z score is > tabulated score at 95% confidence level.
3. Outliers
’
The stem and leaf plot above shows no cases of outliers although the data seems to be slightly
skewed to the right with 1 being the smallest and 9 the largest value. 3 is the most occurring
number within the dataset.

ASSIGNMENTS 3
In the boxplot above, there are no cases of outliers, which are usually shown as dots exceeding
both caps of the plot. From the same plot, we can ascertain that the median number of gifts a
teacher gets from the students is 4.
4. Normality
For normality test, the null hypothesis for both normality tests is that their data is normally
distributed.
According to the normality test results in the table above for the data, the p-values > 0.05, hence
concluding that the data is normally distributed. This is also affirmed by the histogram below,
whose normal curve describes normality.

ASSIGNMENTS 4
5. Mean and Standard Deviation
Statistics
Sample size 20
Mean 4.7
Standard Deviation 2.618
The mean of a number of gifts for the sample was 4.7 with a standard deviation of 2.618.
6. Calculating 2 Score
Z score= sample mean− population mean
standard deviation
¿ 4.7−4
2 =0.35
7. Decision
We fail to reject the null hypothesis
8. Results
The critical value at 5% significance level is 1.96 which is greater than our calculated value. We
conclude that the calculated critical value lies within the acceptance region. Therefore, we
conclude that the number of gifts a teacher receives from students per year at NSW is not
different from 4.

ASSIGNMENTS 5
Tutorial 5 Assessable Exercise
1. Hypothesis
Null hypothesis: The average time a student spends to realize that the language being spoken to
is not English = 60 seconds
Alternative hypothesis: The average time a student spends to realize that the language being
spoken to is not English > 60 seconds
2. Outliers
The stem-and-leaf plot shows that there are no outliers. The minimum value is 45 seconds and a
maximum of 136. The minimum and maximum values are also represented by the boxplot caps –
which shows that there are no outliers.

ASSIGNMENTS 6
According to the boxplot above, the data for the time a student finds out they are not spoken to in
English is approximately normal.
3. How to deal with outliers
If I found outliers in the data set I could decide to transform the data to reduce the variation. For
instance, I can decide to use natural log, hence reducing the effect of the outliers in the analysis.
In some cases, I can decide to remove them in cases where the sample size will not be at risk or
not much information will be lost from the dataset.
4. Normality test

ASSIGNMENTS 7
The normality test assumes that the data is normally distributed. Both tests (Kolmogorov-
Smirnov and Shapiro-Wilk) returns p-value which is greater the significance level (0.05, hence
failing to reject the null hypothesis and conclude that the data is normally distributed.
5. The reason for performing a one-sample t-test
The data was collected for single observation from each individual which is a continuous
variable. There is prior information about the populations time spend to note that the language
being spoken is not English. Therefore, this meets the minimum requirements of performing the
one sample t-test to check whether there is a difference between the known information and what
is observed.
6. Are assumptions of the test met
Yes
- The data is continuous
- The data is independent
- The data is normally distributed
7. The test results
According to the results, the p-value < 0.001, hence rejecting the null hypothesis.
8. Conclusion (5% significance level)
We conclude that the average time for the student to notice that the language they are being
spoken to is not English is greater than 60 seconds.