Estimating the Mean Running Time of Family Movies
The running times of 200 family movies are grouped into intervals with corresponding frequencies. To estimate the mean running time:
Step 1: Calculate Mid-Interval Values
- For each class interval, calculate the midpoint (mid-interval value). For example:
- 70 to 80 minutes: (70 + 80) / 2 = 75
- 80 to 85 minutes: (80 + 85) / 2 = 82.5
- 85 to 90 minutes: 87.5
- 90 to 95 minutes: 92.5
- 95 to 105 minutes: 100
Step 2: Use Frequency Data
- Multiply each mid-interval value by its frequency:
- 75 * 11 = 825
- 82.5 * 51 = 4207.5
- 87.5 * 68 = 5950
- 92.5 * 47 = 4347.5
- 100 * 23 = 2300
Step 3: Calculate the Mean
- Sum of (mid-interval value × frequency) = 825 + 4207.5 + 5950 + 4347.5 + 2300 = 17630
- Total frequency = 200
- Mean running time = 17630 / 200 = 88.15 minutes
Estimating the Interquartile Range (IQR) Using Cumulative Frequency Graph
Step 1: Identify Quartiles
- Total movies = 200
- Q1 (25th percentile) = 0.25 × 200 = 50th movie
- Q3 (75th percentile) = 0.75 × 200 = 150th movie
Step 2: Estimate Q1 and Q3 from Graph
- Q3 corresponds to 91.5 minutes
- Q1 corresponds to 84 minutes
Step 3: Calculate IQR
- IQR = Q3 - Q1 = 91.5 - 84 = 7.5 minutes
Outlier Detection for Starfield Movie
- Starfield running time = 100 minutes
- Calculate upper outlier boundary:
- Upper boundary = Q3 + 1.5 × IQR = 91.5 + 1.5 × 7.5 = 102.75 minutes
- Since 100 < 102.75, Starfield is not an outlier.
Chi-Square Goodness of Fit Test for Normality
Hypotheses
- H0: Running times follow a normal distribution with mean 88 and standard deviation 6.75
- H1: Running times do not follow this normal distribution
Expected Frequencies Calculation
- Use normal cumulative distribution function (CDF) to find probabilities for intervals
- Multiply probabilities by total frequency (200) to get expected frequencies
- Example: For interval 85 to 90 minutes,
- Probability = normalCDF(85, 90, 88, 6.75)
- Expected frequency = Probability × 200 = 57.6
Chi-Square Test Execution
- Observed and expected frequencies entered into calculator
- Degrees of freedom = number of intervals - 1 = 4
- Test statistic χ2 = 12.1
- P-value = 0.0165
Conclusion
- Since p-value (0.0165) < significance level (0.05), reject H0
- Conclusion: The running times of family movies do not follow a normal distribution with mean 88 and standard deviation 6.75
This analysis provides a comprehensive approach to summarizing running time data, estimating central tendency and spread, detecting outliers, and statistically testing distribution assumptions for family movie durations. For further insights on statistical methods, you may find the following resources helpful:
The running time t of 200 family movies are recorded in the following table. Write down the mid interval value of t
greater than or equal to 70 less than 80. Mid interval value equals 70 + 80 over
2. So we get 75. Calculate an estimate of the mean running
time of the 200 movies. First of all, we need to figure out mid interval value.
75 80 + 85 over 2 82.5 87.5
92.5 100 Then we need to put this into calculator to figure out the mean. Since
this is a frequency so only have one variable we need to go to one variable stats click
on one then no we will go to four name column one as a
P then column B as the frequency. So, FR. After you put in all the
values, click on an empty cell menu for statistics. One stats
calculations. This is a one variable stats because the second column is a frequency.
For one variable stats, number of leads is always one. Then enter. Right arrow button to get P as a
X1 list. For frequency list, it's
a FR. Then enter. So we get this xbar which is mean is
88 15 since this is two marks. So we need to show the work. Xar
equals 75 * 11 82.5 *
51 87.5 * 68 92.5
* 47 100 * 23 all over frequency add all these
together equals 200 100 equals
88.15. This table is used to create the following cumulative frequency graph. Use the cumulative frequency
curve to estimate the inter quartile range. Intercortile range
IQR equals Q3 minus Q1. Q3 is 75%. So
75% times cumulative frequency total is
200 equals 150. there at 150 then going
down. Then you need to check the unit for running
time from 70 to 75. We count how many small units here? 75 - 70 over
10 equals 0.5. So 90
plus we have three units of 0.5. 90 + 3 * 0.5 is 91.5. For Q1, the first quartile equals
25%. So 25% times the total
number 200 equals 50. So get this 50. Draw this horizontal line. Then
going down 80 plus all together is a
8 each is a 0.5. So we get 84 equals
91.5 minus 84 equals 7.5. Starfield is a movie in the data set and
its running time is 100 minutes. Use your answer to part B to estimate whether Starfield's running time is an
outlier for this data. Justify your answer. We have a lower outlier.
We will use the closest quartile
minus 1.5 IQR
higher outlier closest quartile is Q3 add
1.5 IQR we will find the higher outline fire
equals Q3 is 91.5 + 1.5 * 7.5 =
102.75. Since 100 less than 102.75, it's
not an outlier. It's believed that the running times of family movies follow a normal
distribution with the mean 88 minutes and a standard deviation 6.75 minutes. We can write down this
normal distribution n with 88 6.75 squared.
is uh decided to perform a kai square goodness of fit test on the data to determine whether this sample of 200
movies could have possibly been drawn from an underlying distribution
N8 6.75 squared. Write down the no and alternative hypothesis for this test. We
want to test the running times of family movies follow a normal
distribution with mean 88 minutes and a standard deviation
6.75 minutes. This is the HO. H1 will be opposite. H1 the running
times of family movies don't follow a normal distribution. N 88
6.75 squared. As a part of test, the following table is created. Find the
value of a and the value of b. We will work out this expected frequency. We will work out this expected frequency
based on the assumption. Ho is true. So we
assume observed the frequency follows a normal distribution with 88 as the mean
6.75 as the standard deviation. So this a equals probability times
n. n is given 200 movies. So a
= 200. Probability we will use normal
CDF times 200 equals 85 lower bound 90 is the upper bound. Go
to this uh calculator scratch pad control menu clear the history. Go to manual
five five to normal CDF 85 upper bound
90 mean 88 standard deviation 6.75
enter then times 200 100
57.6 85 90 88
6.75 B equals we will work out B by subtraction from 200 200 - 23.6
6 - 42.1 - 57.6 -
46.7 equals 30. Hence perform the test to a 5% significance level clearly stating
the conclusion in context. We will put observed frequency and expected
frequency to the spread sheet. Let's go to calculator on
one no for spreadsheet
P. Here's a Q. Type in the observed frequency and expected frequency
separately. Go to menu for statistics for stat tests. Seven K square G of
F right arrow button get a P for observed list expected list will be
Q DF we have five data 5 minus one is a four then enter
Enter K² = 12.1 P value 0.0165 DF = 4 K²
= 12.1 P value = 0.0165
0165 DF = 4. We can draw the conclusion now. P value 0.0165. Alpha equals
0.05. Compare which one is greater? This is a less than we know. Please uh reject ho p
value less than alpha reject ho. We got
0.0165 or less than 0.05. Reject ho. So we write down h1.
H1 the running times of a family movies don't follow a normal distribution and 88
6.75 squared. The running times of family movies do not follow a normal distribution with
88 as the mean, 6.75 as the standard deviation.
Heads up!
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