Display each set of data in a stem-and-leaf plot.
1 answer:
The stem-and-leaf plot for each set of data can be displayed as shown in the images attached below (see attachment).
<h3>What is a Stem-and-leaf Plot?</h3>A stem-and-leaf plot is used to display a data distribution, for example, 25 and 26 can be displayed as:
2 | 5, 6.
Thus, the stem-and-leaf plot for each set of data can be displayed as shown in the images attached below (see attachment).
Learn more about stem-and-leaf plot on:
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Answer:
Step-by-step explanation:
Given that:
A study is conducted and the study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands.
a)
Thus; we formulate the null and the alternative hypotheses as follows:
The proportion is ; = 0.64
Null hypothesis:
The Null hypothesis states that there is no evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.
Alternative hypothesis:
The Alternative hypothesis states that there is evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.
b)
The proportion of the population = 0.64
The sample size n = 100
The number of shoppers = 52 stating that the supermarket brand was as good as the national brand
The sample proportion
The z - value is calculated by the formula:
z = -2.50
Since z = -2.50
the p-value = 2P(Z ≤ -2.50)
p-value = 2 × 0.0062
The p-value = 0.0124
c)
At significance level, ∝ = 0.05 ; The p-value = 0.0124
According to the rejection rule, if p-value is less than 0.05 then we will reject null hypothesis at ∝ = 0.05
Hence, the p-value =0.0124 < ∝ (=0.05)
According to the reject rule; reject null hypothesis.
Conclusion: There is no evidence at all that the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%.
d) we know that the significance level of ∝ = 0.05
The value is obtained as:
Now; to determine Z ; we locate the probability value of 0.05 form the table of standard normal distribution. Then we proceed to the left until the first column is reached and the value is 1.90. Also , we move upward until we reach the top and the value is 0.06. Now; the intersection of the row and column results the area to the left of z
This implies that :
The critical value for left tail is -1.96 and the critical value for right tail is 1.96.
Conclusion:
The critical value is -1.96 and the value of test statistic is - 2.50. Here, we can see that the value of test statistic is lesser than the critical value. Hence, we can be concluded that there is evidence that reject the null hypothesis.
Therefore, Yes, the national brand ketchup manufacturer is pleased with the conclusion.
Answer:
There was a 1.93% probability of having multiples (twins or more) in 1980.
Step-by-step explanation:
There were 3,612,258 births in the US in 1980.
Of those,
68339 + 1337 = 69676 were multiplies.
So
What was the probability of having multiples (twins or more) in 1980?
There was a 1.93% probability of having multiples (twins or more) in 1980.