Question

In: Statistics and Probability

A hunter claims that the average deer is more than 200 pounds. To test his claim,...

A hunter claims that the average deer is more than 200 pounds. To test his claim, a random sample of 20 deer were weighed. The data are listed here. Do these data allow us to infer at the 5% significance level that the hunter’s claim is true? 160 230 180 410 220 180 230 190 220 150 180 350 160 150 170 190 230 150 160 260

Solutions

Expert Solution

A hunter claims that the average deer is more than 200 pounds.

The null and alternative hypotheses are,

H0 : μ = 200 pounds

Ha : μ > 200 pounds

sample size (n) = 20

Using Excel we get, sample mean = 208.5 and

sample standard deviation (s) = 67.6115

Test statistic is,

=> Test statistic = t = 0.562

Degrees of freedom = 20 - 1 = 19

t-critical value at significance level of 0.05 with 19 degrees of freedom is, tcrit = 1.729

Since, test statistic = 0.562 < 1.729, we fail to reject the null hypothesis.

Conclusion : There is not sufficient evidence to support the claim that the average deer is more than 200 pounds.


Related Solutions

A nutrition expert claims that the average American is overweight. To test his claim, a random...
A nutrition expert claims that the average American is overweight. To test his claim, a random sample of 26 Americans was selected, and the difference between each person's actual weight and idea weight was calculated. For this data, we have x¯=16.1x¯=16.1 and s=30s=30. Is there sufficient evidence to conclude that the expert's claim is true? Carry out a hypothesis test at a 6% significance level. A. The value of the standardized test statistic: Note: For the next part, your answer...
A diet doctor claims that the average North American is more than 20 pounds overweight. To...
A diet doctor claims that the average North American is more than 20 pounds overweight. To test his claim, a random sample of 20 North Americans was weighed, and the difference between their actual and ideal weights was calculated. The data are listed here. Do these data allow us to infer at the 5% significance level that the doctor’s claim is true? 16 23 18 41 22 18 23 19 22 15 18 35 16 15 17 19 23 15...
A new​ weight-loss program claims that participants will lose an average of more than 10 pounds...
A new​ weight-loss program claims that participants will lose an average of more than 10 pounds after completing it. The accompanying data table shows the weights of eight individuals before and after the program. Complete parts​ (a) through​ (e) below. Person Before After 1 226 203 2 213 195 3 206 195 4 186 173 5 204 195 6 199 181 7 243 232 8 191 193 . a. Perform a hypothesis test using alphaαequals=0.10 to determine if the average...
A diet doctor claims Australians are, on average, overweight by more than 10kg. To test this...
A diet doctor claims Australians are, on average, overweight by more than 10kg. To test this claim, a random sample of 100 Australians were weighed, and the difference between their actual weight and their ideal weight was calculated and recorded. The data are contained in the Excel file Weights.xlsx. Use these data to test the doctor's claim at the 5% level of significance. Excess weight (Kgs) 16.0 4.0 4.0 4.5 11.0 7.0 7.0 16.5 14.5 5.5 16.5 0.5 13.5 26.0...
A health researcher read the claim that  a 200-pound male can burn more than an average of...
A health researcher read the claim that  a 200-pound male can burn more than an average of 524 calories per hour playing tennis with a standard deviation of 45.9 calories. 37 males were randomly selected and the mean number of calories burnt per hour was 534.8. Test at 5% of level of significance whether the claim is correct? (Hint:  Right Tailed Test) State Null and Alternate Hypotheses. Apply test. Write your conclusion using critical value approach and p value approach. Construct a...
On average, they claim that a 1.69 oz bag will contain more than 54 candies. Test...
On average, they claim that a 1.69 oz bag will contain more than 54 candies. Test this claim (µ > 54) at the 0.01 significance (σ unknown). Blue Orange Green Yellow Red Brown Total Number of Candies in Bag 6 17 10 8 10 7 58 8 8 11 12 9 10 58 8 13 14 4 12 7 58 7 13 10 7 14 7 58 12 13 4 13 5 10 57 13 8 12 13 1 10...
to test the claim that the resistance of electric wire can be reduced by more than...
to test the claim that the resistance of electric wire can be reduced by more than 0.050 ohm by alloying, thirty two (32) values were obtained for standard wire, and the sample average calculated was equal to 0.136 ohm. in addition, thirty two (32) values were obtained for alloyed wire, and the sample average calculated was equal to 0.083 ohm. the standard deviation resistance for standard wire is known to be 0.004 ohm, whereas the standard deviation resistance for alloyed...
A politician claims that the mean salary for managers in his state is more than the...
A politician claims that the mean salary for managers in his state is more than the national? mean, ?$80,000. Assume the population is normally distributed and the   population standard deviation is ?$7400. The salaries? (in dollars) for a random sample of 30 managers in the state are listed. At alpha=0.05?, is there enough evidence to support the? claim? alpha= Hypothesis (use words if needed) Ho: H1: test statistic= p-value= Decision= Interpretation 74,740 82,369 92,359 81,979 95,147 81,132 92,618 75,113 91,420...
a politician claims that the mean salary for managers in his State is more than the...
a politician claims that the mean salary for managers in his State is more than the national mean, $82,000.00. Assume the population is normally distributed and the population deviation is $9500.00. The salaries(in dollars) for a random sample of 30 managers in the state are listed. At a=0.07, is there enough evidence to support the claim? the dollar amounts are:$73,747.00, $87,299.00, $77,242.00, $72,623.00, $80,753.00, $83,263.00, $96,365.00, $93,988.00, $87,831.00, $97,484.00, $74,216.00, $71,883.00, $83,884.00, $89,409.00, $78,373.00, $75,138.00, $83,274.00, $91,670.00, $92,903.00, $74,575.00, $77,973.00,...
A survey claims that the average cost of a hotel room in Atlanta is more than...
A survey claims that the average cost of a hotel room in Atlanta is more than $125.50 per night. To test this claim, a researcher selects a random sample of 55 hotel rooms and finds that the mean cost per night is $130.75. The standard deviation of the population, σ, is known to be $2.50. At a = 0.05, is there enough evidence to support the claim? Use the P-Value Method of Testing. In your work space below, you will...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT