document.write( "Question 1174400: In a certain food stall, 278 out of 500 randomly selected consumers indicate their preference for a new kind of food combination. Use a 99% confidence interval to estimate the true proportion p who like the new food combination.
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Algebra.Com's Answer #850668 by CPhill(1959)\"\" \"About 
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**1. State the Hypotheses:**\r
\n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H0):** The sample is from a population with a mean height of 171.20 cm (µ = 171.20).
\n" ); document.write( "* **Alternative Hypothesis (H1):** The sample is not from a population with a mean height of 171.20 cm (µ ≠ 171.20).\r
\n" ); document.write( "\n" ); document.write( "**2. Determine the Significance Level:**\r
\n" ); document.write( "\n" ); document.write( "* Alpha (α) = 0.05\r
\n" ); document.write( "\n" ); document.write( "**3. Calculate the Test Statistic:**\r
\n" ); document.write( "\n" ); document.write( "* Since we have a large sample size (n = 400) and know the population standard deviation, we'll use a z-test.
\n" ); document.write( "* z = (sample mean - population mean) / (population standard deviation / √sample size)
\n" ); document.write( "* z = (171.40 - 171.20) / (3.3 / √400) = 0.20 / 0.165 ≈ 1.21\r
\n" ); document.write( "\n" ); document.write( "**4. Determine the Critical Value:**\r
\n" ); document.write( "\n" ); document.write( "* For a two-tailed test at α = 0.05, the critical z-value is ±1.96.\r
\n" ); document.write( "\n" ); document.write( "**5. Compare the Test Statistic to the Critical Value:**\r
\n" ); document.write( "\n" ); document.write( "* Our calculated z-value (1.21) falls within the range of -1.96 to +1.96.\r
\n" ); document.write( "\n" ); document.write( "**6. Make a Decision:**\r
\n" ); document.write( "\n" ); document.write( "* Since the test statistic falls within the acceptance region, we **fail to reject the null hypothesis**.\r
\n" ); document.write( "\n" ); document.write( "**7. Conclusion:**\r
\n" ); document.write( "\n" ); document.write( "* At a 5% level of significance, there is not enough evidence to conclude that the sample of 400 males with a mean height of 171.40 cm is not from a population with a mean height of 171.20 cm and a standard deviation of 3.3. Therefore, **it can be reasonably regarded as a sample from that population.**
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