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A psychologist claims that the mean age at which children startwalking is 10 months. Carol wanted to check if this claim is true.She took a random sample of 20 chil...

Question

A psychologist claims that the mean age at which children startwalking is 10 months. Carol wanted to check if this claim is true.She took a random sample of 20 children and found that the mean ageat which these children started walking was 12 months. It is knownthat the ages at which all children start walking are approximatelynormally distributed with a standard deviation 1.2 months. Testthat the mean age at which all children start walking is differentfrom 10 months. What the is observed value

A psychologist claims that the mean age at which children start walking is 10 months. Carol wanted to check if this claim is true. She took a random sample of 20 children and found that the mean age at which these children started walking was 12 months. It is known that the ages at which all children start walking are approximately normally distributed with a standard deviation 1.2 months. Test that the mean age at which all children start walking is different from 10 months. What the is observed value of the test statistic? What is your decision regarding the null hypothesis? Use the significance level as 5%. *



Answers


The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers conducted by Essential Baby.
(TABLE CAN'T COPY) (a) Because the sample size is small, we must verify that the data come from a population that is normally distributed and that the sample size does not contain any outliers. The normal probability plot and box plot are shown next. Are the conditions for constructing a confidence interval about the mean satisfied? (FIGURE CAN'T COPY) (b) Construct a $95 \%$ confidence interval for the mean age at which a baby first crawls. (c) What could be done to increase the accuracy of the interval without changing the level of confidence?

Here for the solution statement. Donald Hypothesis. That note can be stated as soon below that age, not too mu is equal to 2.5. That means the mainland of jail time has not increased from 2.5 years now for the next step. And the alternative hypothesis that it's at note can be stated as shown below that edge. Not, um, you is greater than 2.5. That means the mean length of the jail time has increased from 2.5. Yes. So this is the explanation. Step by step. Please go through this.

Problem wants us to find the Nolan alternative hypothesis in this study. So just as a housekeeping note, normally in statistics, you right mean as the letter mule. It kind of looks like a you. And I'm gonna be using this letter mule to represent mean, just because that's the most common notation now to write no on alternative hypothesis for no, we write h not. And then for alternative, we write h A. So how do we know which one is which? Well, your alternative hypothesis is usually what the researchers trying to find. Or it means there's a difference between two groups or a sample in a population off annoy pop assists usually means there is no difference. There is no change or it's whatever the researcher wasn't going to conclude. So on this study yeah, they want to test if it is actually higher. So this means all our our alternative hypothesis is going to be. That mule is higher than 2.5, Since that's what the original claim waas now are. No hypothesis is usually the opposite. However, you could write this one in two ways. You could put mule is less than or equal to 2.5 or you could put mule is equal to 2.5. That's because in this problem, we're testing if it's higher. So the alternative would be that the new sample isn't higher, so you could interpret that as meaning. It's the same or it's the same or less than. If we were to actually do this whole problem out, it actually would not make a difference, so either of these answers would be acceptable and we are done.

When we going to test if there is enough technique Enough evidence that the new one baby has Irish weight loss off seven houses Given Thesiger difficult level is out by co 20 poisoner want. So they stayed. The hypothesis is new equal to seven forage not and mu is less than seven for each one. So this is the claim and since this left till test, the critical value Z will be a negative number for Alfa equal to zero poisoner one which is negative 2.33 and then for the test value we have 6.5 minus seven Over one point made was were about 30 to this. Give me native one y 57 And so look at this graph We have ah, significant level Alfa equals 0.1 and then the critical value is navy of 2.33 then our test value is nearly 1.57 is outside of the critical region. So at this region we not reject, it's not and we conclude that there is not enough evidence to support the claim. A significant level Alfa equal to 0.1


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