document.write( "Question 1170718: It is known that the weights of apples from a farm are normally distributed. In order to estimate the mean weight, a random sample of 150 apples is considered and the sample mean and population standard deviation are 6 kg and 0.8 kg respectively.\r
\n" ); document.write( "\n" ); document.write( "(a) Construct a 95% confidence interval estimate for the population mean weight of apples.
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\n" ); document.write( "(b) The researcher suggests doing the study again so that 98% confidence interval estimate for the population mean weight of apples is (5.8835,6.1165) kg. How large should the sample size be?
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Algebra.Com's Answer #851105 by CPhill(1959)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "**Given:**\r
\n" ); document.write( "\n" ); document.write( "* Sample size (n) = 150
\n" ); document.write( "* Sample mean (x̄) = 6 kg
\n" ); document.write( "* Population standard deviation (σ) = 0.8 kg\r
\n" ); document.write( "\n" ); document.write( "**(a) Construct a 95% Confidence Interval**\r
\n" ); document.write( "\n" ); document.write( "1. **Find the critical z-value (z_c):**
\n" ); document.write( " * For a 95% confidence interval, z_c = 1.96 (from a z-table or calculator).\r
\n" ); document.write( "\n" ); document.write( "2. **Calculate the margin of error (E):**
\n" ); document.write( " * E = z_c * (σ / √n)
\n" ); document.write( " * E = 1.96 * (0.8 / √150)
\n" ); document.write( " * E ≈ 1.96 * (0.8 / 12.247)
\n" ); document.write( " * E ≈ 1.96 * 0.0653
\n" ); document.write( " * E ≈ 0.128 kg\r
\n" ); document.write( "\n" ); document.write( "3. **Construct the confidence interval:**
\n" ); document.write( " * Confidence Interval = x̄ ± E
\n" ); document.write( " * Confidence Interval = 6 ± 0.128
\n" ); document.write( " * Confidence Interval = (6 - 0.128, 6 + 0.128)
\n" ); document.write( " * Confidence Interval = (5.872, 6.128) kg\r
\n" ); document.write( "\n" ); document.write( "4. **Answer:**
\n" ); document.write( " * The 95% confidence interval for the population mean weight of apples is (5.872, 6.128) kg.\r
\n" ); document.write( "\n" ); document.write( "**(b) Find the sample size for a 98% Confidence Interval**\r
\n" ); document.write( "\n" ); document.write( "1. **Given Confidence Interval:**
\n" ); document.write( " * (5.8835, 6.1165) kg\r
\n" ); document.write( "\n" ); document.write( "2. **Calculate the margin of error (E):**
\n" ); document.write( " * E = (Upper Bound - Lower Bound) / 2
\n" ); document.write( " * E = (6.1165 - 5.8835) / 2
\n" ); document.write( " * E = 0.233 / 2
\n" ); document.write( " * E = 0.1165 kg\r
\n" ); document.write( "\n" ); document.write( "3. **Find the critical z-value (z_c):**
\n" ); document.write( " * For a 98% confidence interval, z_c ≈ 2.33 (from a z-table or calculator).\r
\n" ); document.write( "\n" ); document.write( "4. **Use the margin of error formula to solve for n:**
\n" ); document.write( " * E = z_c * (σ / √n)
\n" ); document.write( " * 0.1165 = 2.33 * (0.8 / √n)
\n" ); document.write( " * √n = (2.33 * 0.8) / 0.1165
\n" ); document.write( " * √n = 1.864 / 0.1165
\n" ); document.write( " * √n ≈ 16.00
\n" ); document.write( " * n = (16.00)²
\n" ); document.write( " * n = 256\r
\n" ); document.write( "\n" ); document.write( "5. **Answer:**
\n" ); document.write( " * The sample size should be 256 apples.
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