Hi, Please answer all parts.- Can normality be assumed?Last parts, the options are:- P Value : greater than OR Less than- Null hypothesis: Fails to be rejected OR Is rejected- Mean GPA: Sufficient OR Insufficient ### Hypothesis Testing of Mean Class GPAA school administrator believes that the mean class GPAs for a given course are higher than the preferred mean of 2.75 at a significance level of 0.05. They randomly choose a sample of 50 classes. The objective is to create a histogram, and calculate the sample mean ( $ \bar{x} $ ), the t-statistic ( $ t $ ), and the p-value ( $ p $ ).#### Steps to Follow:1. **Evaluating Normality Assumption:** - From the histogram, determine if the data distribution is approximately normal. Use the dropdown to select "Yes" or "No" based on the histogram.2. **Calculating and Verifying Test Statistics:** - Enter the calculated sample mean $ \bar{x} $ in the provided textbox. - Calculate the t-statistic and enter it in the relevant field. - Determine the p-value and input it accordingly.3. **Decision Rule:** - Compare the p-value with the significance level (0.05). - If the p-value is less than or greater than the significance level, choose between "Accept" or "Reject" the null hypothesis from the dropdown menu.4. **Conclusion:** - Based on the test results, select the appropriate conclusion regarding evidence that the mean GPA is greater than 2.75 from the dropdown menu.#### Input Fields:- $ \bar{x} = \text{Ex: 1.234} $- $ t = \text{} $- $ p = \text{} $#### Evaluation Dropdowns:- From the histogram, can normality be assumed? - [Pick: Yes / No]- Since the $ p $-value is [Pick: less / greater] than the significance level 0.05, the null hypothesis - [Pick: accept / reject].- [Pick: Strong / Weak] evidence exists that the mean GPA is greater than 2.75.#### Submission Controls:- **Check**: Validates the entered values and selections.- **Next**: Proceeds to the next step or question.This structured approach ensures a thorough analysis of the hypothesis test related to mean class GPAs, allowing for detailed educational understanding and practice of statistical methods. ### Frequency Analysis of 'Field1' DataThe following data represents a series of measurements designated as 'Field1':```2.59, 2.43, 2.96, 2.80, 2.94, 3.07, 2.65, 2.79, 2.91, 2.96,2.43, 3.14, 2.78, 2.90, 2.78, 2.52, 2.60, 2.86, 2.88, 2.78,3.09, 2.82, 2.72, 2.98, 2.68, 2.87, 2.86, 2.78, 2.73, 2.62,2.77, 2.90, 3.02, 2.87, 2.63, 2.98, 2.52, 2.91, 2.71, 3.07,2.89, 2.93, 2.64, 2.94, 2.81, 2.85, 2.89, 2.79, 2.86```### Histogram and Cumulative Frequency DistributionA histogram is presented below to showcase the frequency distribution of the 'Field1' values, divided into different intervals or bins. Alongside, a cumulative frequency (CDF) curve is depicted to illustrate the cumulative percentage.#### Description of the Histogram- **X-Axis (Field1 Intervals)**: The range of 'Field1' values is grouped into distinct intervals (or bins): - (2.75, 2.83] - (2.83, 2.91] - (2.91, 2.99] - (2.59, 2.67] - (2.67, 2.75] - (2.51, 2.59] - (2.99, 3.07] - (2.43, 2.51] - (3.07, 3.15]- **Y-Axis (Frequency)**: Reflects the number of occurrences (frequency) of 'Field1' values within each interval. The frequency range varies from 0 to 14

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Description: Hi, Please answer all parts.- Can normality be assumed?Last parts, the options are:- P Value : greater than OR Less than- Null hypothesis: Fails to be rejected OR Is rejected- Mean GPA: Sufficient OR Insufficient ### Hypothesis Testing of Mean Class

Given:n = 50level of significance = 0.05Sample data:…

Answered: A school administrator believes that…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Hi, Please answer all parts.

- Can normality be assumed?

Last parts, the options are:

- P Value : greater than OR Less than

- Null hypothesis: Fails to be rejected OR Is rejected

- Mean GPA: Sufficient OR Insufficient

### Hypothesis Testing of Mean Class GPA

A school administrator believes that the mean class GPAs for a given course are higher than the preferred mean of 2.75 at a significance level of 0.05. They randomly choose a sample of 50 classes. The objective is to create a histogram, and calculate the sample mean ( $ \bar{x} $ ), the t-statistic ( $ t $ ), and the p-value ( $ p $ ).

#### Steps to Follow:

1. **Evaluating Normality Assumption:**
- From the histogram, determine if the data distribution is approximately normal. Use the dropdown to select "Yes" or "No" based on the histogram.

2. **Calculating and Verifying Test Statistics:**
- Enter the calculated sample mean $ \bar{x} $ in the provided textbox.
- Calculate the t-statistic and enter it in the relevant field.
- Determine the p-value and input it accordingly.

3. **Decision Rule:**
- Compare the p-value with the significance level (0.05).
- If the p-value is less than or greater than the significance level, choose between "Accept" or "Reject" the null hypothesis from the dropdown menu.

4. **Conclusion:**
- Based on the test results, select the appropriate conclusion regarding evidence that the mean GPA is greater than 2.75 from the dropdown menu.

#### Input Fields:

- $ \bar{x} = \text{Ex: 1.234} $
- $ t = \text{} $
- $ p = \text{} $

#### Evaluation Dropdowns:

- From the histogram, can normality be assumed?
- [Pick: Yes / No]

- Since the $ p $-value is [Pick: less / greater] than the significance level 0.05, the null hypothesis
- [Pick: accept / reject].

- [Pick: Strong / Weak] evidence exists that the mean GPA is greater than 2.75.

#### Submission Controls:

- **Check**: Validates the entered values and selections.
- **Next**: Proceeds to the next step or question.

This structured approach ensures a thorough analysis of the hypothesis test related to mean class GPAs, allowing for detailed educational understanding and practice of statistical methods.

### Frequency Analysis of 'Field1' Data

The following data represents a series of measurements designated as 'Field1':

```
2.59, 2.43, 2.96, 2.80, 2.94, 3.07, 2.65, 2.79, 2.91, 2.96,
2.43, 3.14, 2.78, 2.90, 2.78, 2.52, 2.60, 2.86, 2.88, 2.78,
3.09, 2.82, 2.72, 2.98, 2.68, 2.87, 2.86, 2.78, 2.73, 2.62,
2.77, 2.90, 3.02, 2.87, 2.63, 2.98, 2.52, 2.91, 2.71, 3.07,
2.89, 2.93, 2.64, 2.94, 2.81, 2.85, 2.89, 2.79, 2.86
```

### Histogram and Cumulative Frequency Distribution

A histogram is presented below to showcase the frequency distribution of the 'Field1' values, divided into different intervals or bins. Alongside, a cumulative frequency (CDF) curve is depicted to illustrate the cumulative percentage.

#### Description of the Histogram

- **X-Axis (Field1 Intervals)**: The range of 'Field1' values is grouped into distinct intervals (or bins):
- (2.75, 2.83]
- (2.83, 2.91]
- (2.91, 2.99]
- (2.59, 2.67]
- (2.67, 2.75]
- (2.51, 2.59]
- (2.99, 3.07]
- (2.43, 2.51]
- (3.07, 3.15]

- **Y-Axis (Frequency)**: Reflects the number of occurrences (frequency) of 'Field1' values within each interval. The frequency range varies from 0 to 14

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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