Question

In: Statistics and Probability

A biologist observes a sample of 25 cells to have a mean dividing time of 30...

A biologist observes a sample of 25 cells to have a mean dividing time of 30 minutes with a standard deviation of 3.2 minutes.

a. Construct a 99% confidence interval for population mean dividing time.

b. A colleague claims that the mean dividing time is 32 minutes, but the biologist believes that it is less than 32 minutes. Carry out a hypothesis test and decide who is right - the biologist or the colleague.

c. State your conclusion in context of the hypothesis test and confidence interval.

d. What is the possible error made? What does the error mean in this problem situation?

Solutions

Expert Solution

a)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   24          
't value='   tα/2=   2.797   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.2/√25=   0.6400          
margin of error , E=t*SE =   2.7969   *   0.6400   =   1.7900
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    30.00   -   1.790041   =   28.2100
Interval Upper Limit = x̅ + E =    30.00   -   1.790041   =   31.7900
99%   confidence interval is (   28.21   < µ <   31.79   )

b)

Ho :   µ =   32  
Ha :   µ <   32   (Left tail test)
          
Level of Significance ,    α =    0.010  
sample std dev ,    s =    3.2000  
Sample Size ,   n =    25  
Sample Mean,    x̅ =   30.0000  
          
degree of freedom=   DF=n-1=   24  
          
Standard Error , SE = s/√n =   3.2/√25=   0.6400  
t-test statistic= (x̅ - µ )/SE =    (30-32)/0.64=   -3.125  
          
critical t value, t* =        -2.492   [Excel formula =t.inv(α/no. of tails,df) ]
          
p-Value   =   0.0023   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value<α, Reject null hypothesis       
Conclusion: There is enough evidence that biologist is correct

c)

  There is enough evidence to reject the null hypothesis that true mean is 32 minutes

there is 99% confidence that true ,mean lies within confidence interval

d)

Type I error could be made

Type I error is concluding that mean is less than 32 but infact it is 32 minutes


Related Solutions

a random sample of 25 people the mean commute time to work is 34.5 min and...
a random sample of 25 people the mean commute time to work is 34.5 min and the standard deviation was 7.3 min. use t distribution to construct an 80% confidence interval for the population mean. what is the margin of error
A magazine claims the mean monthly cost of video games in $30. You Sample 25 houses...
A magazine claims the mean monthly cost of video games in $30. You Sample 25 houses and find a mean of $25.7 and a standard deviation of $5.12. At α=.1, do you have enough evidence to reject the claim
Suppose you have a sample of size 30 with a mean of 48 and a population...
Suppose you have a sample of size 30 with a mean of 48 and a population standard deviation of 13.1. What is the maximal margin of error associated with a 95% confidence interval for the true population mean? In your calculations, use z = 2. Give your answer as a decimal, to two places m =  ounces
A biologist examines 6 geological samples for lead concentration. The mean lead concentration for the sample...
A biologist examines 6 geological samples for lead concentration. The mean lead concentration for the sample data is 0.714 cc/cubic meter with a standard deviation of 0.0126. Determine the 90% confidence interval for the population mean lead concentration. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A biologist examines 6 geological samples for lead concentration. The mean lead concentration for the sample...
A biologist examines 6 geological samples for lead concentration. The mean lead concentration for the sample data is 0.714 cc/cubic meter with a standard deviation of 0.0126. Determine the 90% confidence interval for the population mean lead concentration. Assume the population is approximately normal. Step 2 of 2: Construct the 90% confidence interval. Round your answer to three decimal places.
A biologist examines 23 sedimentary samples for bromide concentration. The mean bromide concentration for the sample...
A biologist examines 23 sedimentary samples for bromide concentration. The mean bromide concentration for the sample data is 0.349 cc/cubic meter with a standard deviation of 0.0527. Determine the 98% confidence interval for the population mean bromide concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A biologist examines 29 seawater samples for iron concentration. The mean iron concentration for the sample...
A biologist examines 29 seawater samples for iron concentration. The mean iron concentration for the sample data is 0.334 cc/cubic meter with a standard deviation of 0.0139. Determine the 99% confidence interval for the population mean iron concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A biologist examines 17 seawater samples for magnesium concentration. The mean magnesium concentration for the sample...
A biologist examines 17 seawater samples for magnesium concentration. The mean magnesium concentration for the sample data is 0.252 cc/cubic meter with a standard deviation of 0.074. Determine the 80% confidence interval for the population mean magnesium concentration. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A simple random sample of 30 items resulted in a sample mean of 30. The population...
A simple random sample of 30 items resulted in a sample mean of 30. The population standard deviation is 15. a. Compute the 95% confidence interval for the population mean. Round your answers to one decimal place. ( ,  ) b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places. ( ,  )
The television habits of 30 randomly selected children were observed. The sample mean of time watching...
The television habits of 30 randomly selected children were observed. The sample mean of time watching TV was found to be 48.2 hours per week, with a standard deviation of 12.4 hours per week. Test the claim that the children on average spend more than 47 hours per week watching TV. sig. level .05
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT