Question

Since previous studies have reported that elite athletes are often deficient in their nutritional intake (for example, total calories, carbohydrates, protein), a group of researchers decided to evaluate Canadian high-performance athletes. A total of n = 324 athletes from eight Canadian sports centers participated in the study. One reported finding was that the average caloric intake among the n = 201 women was 2403.7 kilocalories per day (kcal/d). The recommended amount is 2811.5 kcal/d.

For one part of the study, n = 114 male athletes from eight Canadian sports centers were surveyed. Their average caloric intake was 3078.0 kilocalories per day (kcal/d) with a standard deviation of 987.0. The recommended amount is 3421.4. Is there evidence that Canadian high-performance male athletes are deficient in their caloric intake?

(a) State the appropriate H₀.

i) H₀: μ > 3421.4

ii) H₀: μ ≠ 3421.4

iii) H₀: μ = 3421.4

iv) H₀: μ ≤ 3421.4

v) H₀: μ < 3421.4

State the appropriate H_a.

i) H_a: μ < 3421.4

ii) H_a: μ ≠ 3421.4

iii) H_a: μ > 3421.4

iv) H_a: μ = 3421.4

v) H_a: μ ≥ 3421.4

(b) Carry out the test. (Round your answer for t to three decimal places.)

t =

Give the degrees of freedom.
df=
Give the P-value. (Round your answer to four decimal places.)
P-value=

State your conclusion.

i) We do not have sufficient evidence to conclude that Canadian high-performance male athletes are deficient in their calorie intake.

ii) We have sufficient evidence to conclude that Canadian high-performance male athletes are deficient in their calorie intake.    

(c) Construct a 95% confidence interval for the daily average deficiency in caloric intake. (Round your answers to one decimal place.)

( , ) kcal/day

Answer 1

Sample size n = 114

Sample mean = 3078.0 kcal/d

Sample standard deviation = s = 987.0

a ) Hypothesis :

null hypothesis

Alternative hypothesis

Left tailed test

b) test statistic :

Degrees of freedom = n - 1 = 114 - 1 = 113

P-value : P-value for this left tailed test is ,

P-value = P( t < test statistic ) = P( t < -3.715 )

Using Excel ,   =T.DIST( t , df , cumulative )

P( t < -3.715 ) = T.DIST( -3.715 , 113 , 1 ) = 0.0002

P-value = 0.0002

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level

Let ,   = 0.05

It is observed that p-value is less than   = 0.05

So reject null hypothesis

Conclusion :

We have sufficient evidence to conclude that Canadian high-performance male athletes are deficient in their calorie intake.