Shana is at the grocery store comparing two brands of juice. Brand A costs $3.84 for a 16-ounce bottle. Brand B costs $4.50 for
a 25- ounce bottle.
To Compare the costs, Shana must compare prices for equal amounts of juice. How can she do this?
Complete The Tables.
Brand A:
Ounces: 16, 8, 4, 2, 1
Prices for 16 and 8: 3.84, 1.92
1. What are the prices for 4,2, 1 if they were divided by 2?
Brand B:
Ounces: 25, 5, 1
Price for 25: 4.50
2. What are the prices for 5, 1 if they were divided by 5?
3. Brand A costs how many per ounce?
4. Brand B costs how many per ounce?
5. Describe another method to compare the costs.
Solve as many questions as you can. Thanks! :D
To Compare the costs, Shana must compare prices for equal amounts of juice. How can she do this?
Complete The Tables.
Brand A:
Ounces: 16, 8, 4, 2, 1
Prices for 16 and 8: 3.84, 1.92
1. What are the prices for 4,2, 1 if they were divided by 2?
Brand B:
Ounces: 25, 5, 1
Price for 25: 4.50
2. What are the prices for 5, 1 if they were divided by 5?
3. Brand A costs how many per ounce?
4. Brand B costs how many per ounce?
5. Describe another method to compare the costs.
Solve as many questions as you can. Thanks! :D
1 answer:
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Answer:
The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.
Step-by-step explanation:
This is a hypothesis test for the difference between proportions.
The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.
The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.
The difference between proportions is (p1-p2)=-0.02708.
The pooled proportion, needed to calculate the standard error, is:
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the z-statistic as:
This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):
As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.