Question
In: Statistics and Probability
Use the chi-square test for independence and the observed frequencies below to determine if the frequency...
- Use the chi-square test for independence and the observed frequencies below to determine if the frequency pattern of COVID-19 cases between males and females differs between the states of Nebraska and Wyoming (p < .05).
*Do not carry out steps 1-2 of null hypothesis testing*
|
Male |
Female |
|
|
Nebraska |
989 |
659 |
|
Wyoming |
237 |
193 |
Solutions
Expert Solution
The hypothesis being tested is:
H0: The frequency pattern of COVID-19 cases between males and females does not differs between the states of Nebraska and Wyoming
Ha: The frequency pattern of COVID-19 cases between males and females differs between the states of Nebraska and Wyoming
| Male | Female | Total | ||
| Nebraska | Observed | 989 | 659 | 1648 |
| Expected | 972.30 | 675.70 | 1648.00 | |
| O - E | 16.70 | -16.70 | 0.00 | |
| (O - E)² / E | 0.29 | 0.41 | 0.70 | |
| Wyoming | Observed | 237 | 193 | 430 |
| Expected | 253.70 | 176.30 | 430.00 | |
| O - E | -16.70 | 16.70 | 0.00 | |
| (O - E)² / E | 1.10 | 1.58 | 2.68 | |
| Total | Observed | 1226 | 852 | 2078 |
| Expected | 1226.00 | 852.00 | 2078.00 | |
| O - E | 0.00 | 0.00 | 0.00 | |
| (O - E)² / E | 1.39 | 1.99 | 3.38 | |
| 3.38 | chi-square | |||
| 1 | df | |||
| .0660 | p-value |
The p-value is 0.0660.
Since the p-value (0.0660) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Therefore, we cannot conclude that the frequency pattern of COVID-19 cases between males and females differs between the states of Nebraska and Wyoming.
Thank you! :)
orchestra answered 2 years agoRelated Solutions
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